Giáo trình Toán cao cấp B2
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- . . ˆ ´. Chuong 1. MA TRA. N-D- I.NH THUC (8+4) I. Ma trˆa.n . . . * Cho m, n nguyˆenduong. Ta go.i ma trˆa.nc˜o m × n l`amˆo.tba’ng sˆo´ gˆo`m m × n . . . . sˆo´ thu. cd¯uo. cviˆe´t th`anh m h`ang, n cˆo.t c´oda.ng nhu sau: a1,1 a1,2 a1,n a2,1 a2,2 a2,n (ai,j )m×n = am,1 am,2 am,n . trong d¯´oc´acsˆo´ thu. c ai,j ,i= 1,m,j = 1,n . . . d¯ u o. cgo.il`ac´acphˆa` ntu’ cu’a ma trˆa.n,chı’ sˆo´ i chı’ h`angv`achı’ sˆo´ j chı’ cˆo.tcu’a . phˆa`ntu’ ma trˆa.n. . . . . . . * Ma trˆa.nc˜o1 × n d¯ u o. cgo.il`ama trˆa.n h`ang, ma trˆa.nc˜om × 1d¯uo. cgo.il`ama . . . trˆa.ncˆo.t, ma trˆa.nc˜on × n d¯ u o. cgo.il`ama trˆa.n vuˆongcˆa´p n. . . . *Trˆen ma trˆa.n vuˆongcˆa´p n,d¯u`ong ch´eo gˆo`m c´acphˆa`ntu’ ai,i,i= 1,n . . . . . . . d¯ u o. cgo.il`ad¯ u `o ng ch´eoch´ınh,d¯u`ong ch´eo gˆo`m c´acphˆa`ntu’ ai,n+1−i,i= 1,n . . . . d¯ u o. cgo.il`ad¯ u `o ng ch´eophu. cu’a ma trˆa.n. . . . * Ma trˆa.n vuˆongcˆa´p n c´oc´acphˆa`ntu’ n˘a`m ngo`aid¯u`ong ch´eo ch´ınh d¯ˆ`eub˘a`ng 0, ngh˜ıal`a: ai,j =0, ∀i =6 j . . d¯ u o. cgo.il`ama trˆa.n ch´eo. * Ma trˆa.n ch´eo c´o ai,i =1,i= 1,n . . . d¯ u o. cgo.il`ama trˆa.nd¯onvi. cˆa´p n,k´yhiˆe.u In. . * Ma trˆa.nc˜om × n c´o ai,j =0, ∀i, j : i>j . . d¯ u o. cgo.il`ama trˆa.nbˆa.c thang. . . . . * Ma trˆa.nc˜om × n c´oc´acphˆa`ntu’ d¯` ˆe ub˘a`ng 0 d¯uo. cgo.il`ama trˆa.n khˆong,k´y hiˆe.u0m,n. *Tago.i ma trˆa.n chuyˆe’nvi. a1,1 a2,1 am,1 T a1,2 a2,2 am,2 A =(aj,i)n×m = a1,n a2,n am,n Typeset by AMS-TEX
- 2 cu’a ma trˆa.n a1,1 a1,2 a1,n a2,1 a2,2 a2,n A =(ai,j )m×n = am,1 am,2 am,n . . . l`ama trˆa.n c´od¯uo. ct`uA b˘a`ng c´ach chuyˆe’n h`angth`anhcˆo.t, cˆo.t th`anhh`ang. . . . * Hai ma trˆa.nc`ung c˜o (ai,j )m×n v`a(bi,j )m×n d¯ u o. cgo.il`ab˘a`ng nhau nˆe´u c´acphˆa`n . . . tu’ o’ t`ung vi. tr´ıd¯ˆ`eub˘a`ng nhau: ai,j = bi,j , ∀i = 1,m,∀j = 1,n. . . +Tˆo’ng (hiˆe.u) cu’a hai ma trˆa.n c`ungc˜o m × n l`amˆo.t ma trˆa.nc˜om × n, trong d¯´o . . . . . . phˆa`ntu’ cu’a ma trˆa.ntˆo’ng (hiˆe.u) l`atˆo’ng (hiˆe.u) c´acphˆa`ntu’ o’ vi. tr´ıtuong ´ung: (ci,j)m×n =(ai,j )m×n ± (bi,j )m×n v´o.i ci,j = ai,j ± bi,j , ∀i = 1,m,∀j = 1,n. . . . . . . +T´ıch vˆohu´ong cu’asˆo´ thu. c α v´oi ma trˆa.nc˜om × n l`ama trˆa.nc˜om × n, trong d¯´o . . . . . . . mˆo˜i phˆa`ntu’ l`at´ıch cu’a α v´oi phˆa`ntu’ o’ vi. tr´ıtuong ´ung cu’a ma trˆa.n ban d¯ˆa`u: (ci,j )m×n = α.(ai,j )m×n v´o.i ci,j = α.bi,j , ∀i = 1,m,∀j = 1,n. . . . +T´ıch vˆohu´ong c´ot´ınhphˆanbˆo´ v´oi ph´epcˆo.ng c´acma trˆa.n: α.(A+B)=α.A+α.B, . . v´oi ph´epcˆo.ng c´achˆe. sˆo´:(α + β).A = α.A + β.B, c´ot´ınhkˆe´tho. p: α.(β · A)=(α.β) · A. +T´ıch cu’a hai ma trˆa.n A =(ai,j )m×n v`a B =(bj,k)n×q l`ama trˆa.n C = A × B =(ci,k)m×q , v´o.i n ci,k = Xai,j bj,k, ∀i = 1,m,∀k = 1,q. j=1 V´ıdu 132 13 1.1+3.1+2.31.3 − 3.1+2.2 10 4 247 × 1 −1 = 2.1+4.1+7.32.3 − 4.1+7.2 = 27 16 356 32 3.1+5.1+6.33.3 − 5.1+6.2 26 16
- 3 . +Ph´ep nhˆanhai ma trˆa.n c´ot´ınhkˆe´tho. p: A × (B × C)=(A × B) × C, t´ınhphˆan . phˆo´id¯ˆo´iv´oi ph´epcˆo.ng: A × (B + C)=A × B + A × C;(A + B) × C = A × C + B × C. Ngo`aira, nˆe´u A c´oc˜o. m × n,th`ı A × In = Im × A = A. . II. D- .inh th´uc * Cho E = {1, 2, 3, ,n}.Tago.i ho´anvi. cu’atˆa.p E l`amˆo.t song ´anh f : E → E, k´yhiˆe.u 12 n f : f(1) f(2) f(n) hay (f(1),f(2), ,f(n)) (c´otˆa´tca’ n! ho´anvi. kh´acnhau). . V´ıdu Cho E = {1, 2, 3}. Anh´ xa. f : E → E x´acd¯i.nh bo’ i: f(1) = 1,f(2) = 3,f(3) = 2 l`amˆo.t ho´anvi. cu’a E,k´yhiˆe.ul`a 123 132 ho˘a.c (1, 3, 2). * Cho mˆo.t ho´anvi. 12 n f : f(1) f(2) f(n) . . ta th`anhlˆa.p c´acc˘a.pth´utu. (f(i),f(j)), ∀i =6 j, 2 . . . . . s˜ec´oCn c˘a.pth´utu. nhu thˆe´;mˆo.tc˘a.p(f(i),f(j)) d¯uo. cgo.il`anghi.ch thˆe´ nˆe´u (i − j)(f(i) − f(j)) < 0. 2 . . Go.i N(f) l`asˆo´ c´acnghi.ch thˆe´ cu’a ho´anvi. f (c´otrong Cn c˘a.pth´utu. trˆen). V´ıdu T`ım sˆo´ nghi.ch thˆe´ cu’a ho´anvi. 12345 f : . 32154
- 4 . . . T`u ho´anvi. n`ay, ta c´oc´acc˘a.pth´utu. (3, 2), (3, 1), (3, 5), (3, 4), (2, 1), (2, 5), (2, 4), (1, 5), (1, 4), (5, 4), trong d¯´ota c´oc´acnghi.ch thˆe´: (3, 2), (3, 1), (2, 1), (5, 4), suy ra N(f)=4 - . . . * Cho ma trˆa.n(A)n,n. D.inh th´uccu’a A l`amˆo.tsˆo´ thu. c, k´yhiˆe.u v`ax´acd¯i.nh nhu sau: N(f) det(A)= X (−1) a1,f(1)a2,f(2) an,f(n) f∈Sn . . trong d¯´o Sn l`atˆa.ptˆa´tca’ n! ho`anvi. cu’a n phˆa`ntu’ {1, 2, ,n}.Nhuvˆa.y, d¯i.nh . th´uccu’a ma trˆa.n A l`amˆo.tsˆo´: . +b˘a`ng tˆo’ng d¯a.isˆo´ cu’a n!ha.ng tu’ da.ng a1,f(1)a2,f(2) an,f(n) . . +mˆo˜iha.ng tu’ l`at´ıch cu’a n phˆa`ntu’ ai,j m`amˆo˜i h`ang,mˆo˜icˆo.t pha’ic´omˆo.t . v`achı’ mˆo.t phˆa`ntu’ tham gia v`aot´ıch d¯´o. . . . . +dˆa´ucu’amˆo˜iha.ng tu’ phu. thuˆo.c v`aosˆo´ nghi.ch thˆe´ cu’a ho´anvi. tuong ´ung. . . . . . *Tago.i d¯ i .nh th´uccˆa´p2l`agi´atri. t´ınhd¯uo. ct`uba’ng 2 h`ang,2 cˆo.tnhusau: a1,1 a1,2 = a1,1a2,2 − a2,1a1,2 a2,1 a2,2 . . . . . *Tago.i d¯ i .nh th´uccˆa´p3l`agi´atri. t´ınhd¯uo. ct`uba’ng 3 h`ang,3 cˆo.tnhusau: a1,1 a1,2 a1,3 a2,1 a2,2 a2,3 = a1,1a2,2a3,3 + a2,1a3,2a1,3 + a3,1a1,2a2,3 a a a 3,1 3,2 3,3 − a3,1a2,2a1,3 − a2,1a1,2a3,3 − a1,1a3,2a2,3 . . . +D- ˆe ’ t´ınhnhanh d¯i.nh th´uccˆa´p 3, ta viˆe´tcˆo.tth´unhˆa´t v`ath´u hai tiˆe´p theo v`aobˆen pha’iba’ng n´oitrˆen: a1,1 a1,2 a1,3 a1,1 a1,2 a2,1 a2,2 a2,3 a2,1 a2,2 a3,1 a3,2 a3,3 a3,1 a3,2 . . . . th`ı3 phˆa`ntu’ lˆa´ydˆa´ucˆo.ng l`at´ıch c´acphˆa`ntu’ n˘a`m trˆenc´acd¯u`ong ch´eo song song v´o.id¯u.`o.ng ch´eo ch´ınh, ba phˆa`ntu’. lˆa´ydˆa´utr`u. l`at´ıch c´acphˆa`ntu’. n˘a`m trˆenc´ac . . . . . d¯ u `ong ch´eo song song v´oid¯u`ong ch´eo phu. (quy t˘a´c Serrhus)
- 5 . . . . *Tago.i d¯ i .nh th´uccˆa´p n l`agi´atri. t´ınhd¯uo. ct`uba’ng: a1,1 a1,2 a1,n a2,1 a2,2 a2,n n+1 = a1,1D1 − a2,1D2 + ···+(−1) an,1Dn an,1 an,2 an,n . . . . trong d¯´o Dk l`ad¯i.nh th´uccˆa´p n − 1thud¯uo. ct`uba’ng d¯˜acho b˘a`ng c´ach bo’ cˆo.t th´u. nhˆa´t v`ah`angth´u. k, k = 1,n. V´ıdu 1452 331 452 452 452 0331 =1. 040 − 0. 040 +2. 331 − 0. 331 =14 2040 021 021 021 040 0021 . +D- .inh th´uc khˆongthay d¯ˆo’inˆe´u ta d¯ˆo’i h`angth`anhcˆo.t . . +D- .inh th´ucd¯ˆo’idˆa´unˆe´u ta d¯ˆo’ichˆo˜ hai h`ang(ho˘a.c hai cˆo.t) v´oi nhau . . +D- .inh th´uc c´ohai h`ang(ho˘a.c hai cˆo.t) ty’ lˆe. v´oi nhau nhau th`ıb˘a`ng 0 . . . +Th`uasˆo´ chung cu’amˆo.t h`anghay cˆo.tc´othˆe’ d¯ u a ra ngo`aidˆa´ucu’ad¯i.nh th´uc . . . +D- .inh th´uc khˆongthay d¯ˆo’inˆe´u ta d¯ˆo`ng th`oicˆo.ng v`aoc´acphˆa`ntu’ cu’amˆo.t h`ang . . (hay mˆo.tcˆo.t) n`aod¯´oc´acphˆa`ntu’ cu’amˆo.t h`ang(hay mˆo.tcˆo.t) kh´acnhˆanv´oic`ung mˆo.tsˆo´. . . V´ıdu Gia’iphuong tr`ınh: 11 1 1 11− x 1 1 112− x 1 =0. 11 1 n− x . . . . . D- .inh th´uco’ vˆe´ tr´aicu’aphuong tr`ınhl`ad¯ath´ucbˆa.c n nˆenc´okhˆongqu´a n nghiˆe.m . kh´acnhau. Thay x =0,x =1,x =2, ,x = n − 1 v`aod¯i.nh th´uc, ta luˆonc´ohai . . . . . h`angv´oi c´acphˆa`ntu’ b˘a`ng 1, nˆen d¯i.nh th´ucb˘a`ng 0. Vˆa.yphuong tr`ınhc´o n nghiˆe.m x =0,x=1,x=2, ,x = n − 1. . . *D- .inh th´uccu’a ma trˆa.n vuˆong A =(ai,j )n×n,k´yhiˆe.u det(A) l`ad¯i.nh th´uccˆa´p n cu’aba’ng a1,1 a1,2 a1,n a2,1 a2,2 a2,n an,1 an,2 an,n v`ac´ot´ınhchˆa´t: + det(αA)=αn. det(A) + det(A × B) = det(A). det(B) III. Ma trˆa.n nghi.ch d¯a’o
- 6 . . −1 * Ma trˆa.n A =(ai,j )n×n d¯ u o. cgo.il`ama trˆa.n kha’ nghi.ch nˆe´utˆo`nta.i ma trˆa.n A sao cho: −1 −1 A × A = A × A = In. −1 . . Khi d¯´o,ma trˆa.n A d¯ u o. cgo.il`ama trˆa.n nghi.ch d¯a’o cu’a A. + Ma trˆa.n A kha’ nghi.ch khi v`achı’ khi det A =06 . . * Cho A =(ai,j )m×n.Mˆo.t d¯ i .nh th´uc con cˆa´p k (1 ≤ k ≤ n) cu’a A l`amˆo.td¯i.nh . . th´ucta.o th`anht`u ma trˆa.n A b˘a`ng c´ach bo’ d¯ i m − k h`angv`a n − k cˆo.t. * Cho ma trˆa.n vuˆongcˆa´p n kha’ nghi.ch a1,1 a1,2 a1,n a2,1 a2,2 a2,n A = an,1 an,2 an,n . i+j Phˆa` nb`ud¯a.isˆo´ cu’a phˆa`ntu’ ai,j ,l`asˆo´ Ai,j =(−1) Di,j trong d¯´o Di,j l`ad¯i.nh . . . . . th´uccˆa´p n − 1cu’aba’ng thu d¯uo. ct`uma trˆa.n A b˘a`ng c´ach ga.ch bo’ h`angth´u i v`a . cˆo.tth´uj. + Cho A l`ama trˆa.n vuˆongkha’ nghi.ch cˆa´p n v`a∆ = det A =6 0. Khi d¯´oma trˆa.n . . . nghi.ch d¯a’ocu’a A d¯ u o. c x´acd¯i.nh mˆo.t c´ach duy nhˆa´tbo’ i: −1 1 T A = Ai,j ∆ A1,1 A2,1 Dn,1 1 A A D = 1,2 2,2 n,2 ∆ A1,n A2,n Dn,n V´ıdu Ma trˆa.n nghi.ch d¯a’ocu’a 1 −11 A = 211 112 l`a: 13−2 1 A−1 = −31 1 5 1 −23 v`ı: ∆ = det A = (1)(1)(2)+(2)(1)(1)+(1)(−1)(1)−(1)(1)(1)−(2)(−1)(2)−(1)(1)(1) = 5 =06
- 7 v`a: 11 21 21 A =(−1)1+1 =1;A =(−1)1+2 = −3; A =(−1)1+3 =1; 1,1 12 1,2 12 1,3 11 2+1 −11 2+2 11 2+3 1 −1 A2,1 =(−1) =3;A2,2 =(−1) =1;A2,3 =(−1) = −2 12 12 11 −11 11 1 −1 A =(−1)3+1 = −2; A =(−1)3+2 =1;A =(−1)3+3 =3 3,1 11 3,2 21 3,3 21 +T´ınh chˆa´t: 1 − Cho A kha’ d¯ a’ov`ak =6 0, th`ı:(kA)−1 = A−1 k − Cho A, B c`ung cˆa´p v`akha’ d¯ a’o, th`ı:(A × B)−1 = B−1 × A−1 −1 − Cho A kha’ d¯ a’oth`ıA−1 c˜ung kha’ d¯ a’ov`a A−1 = A Phˆa`n I.4: Ha.ng cu’a ma trˆa.n *Tago.i ha.ng cu’a ma trˆa.n A =(ai,j )m×n,k´yhiˆe.u r(A) l`acˆa´p cao nhˆa´tcu’a c´ac . d¯ i .nh th´uc con kh´ac0 cu’a A. . +Ha.ng cu’a ma trˆa.n0m×n l`a0, ha.ng cu’a ma trˆa.n A =(a)v´oi a =6 0 l`a1. . +Ha.ng cu’a ma trˆa.n khˆongthay d¯ˆo’i qua c´acph´epbiˆe´nd¯ˆo’iso cˆa´p sau d¯ˆay: a. D- ˆo’ichˆo˜ hai h`angho˘a.c hai cˆo.t cho nhau; . b. Nhˆanmˆo.t h`ang(hay mˆo.tcˆo.t) v´oimˆo.tsˆo´ kh´ac0; . c. Cˆo.ng v`aomˆo.t h`ang(hay mˆo.tcˆo.t) v´oimˆo.t h`ang(hay mˆo.tcˆo.t) kh´acnhˆan . v´oimˆo.tsˆo´. . . D- ˆe ’ t`ımha.ng cu’a ma trˆa.n Amtimesn, c´othˆe’ d`ungc´acphuong ph´apsau: . . . . . + Phuo ng ph´aptheo d¯i.nh ngh˜ıa: t´ınhc´acd¯i.nh th´uc con t`u cˆa´p 2 tro’ lˆen.Gia’ . . . su’ ma trˆa.nc´o1d¯i.nh th´uc con cˆa´p r kh´ac0, t´ınhtiˆe´p c´acd¯i.nh th´uccˆa´p r +1,nˆe´u . tˆa´tca’ d¯` ˆe ub˘a`ng 0 th`ıkˆe´t luˆa.nha.ng ma trˆa.nl`ar,nˆe´uc´od¯i.nh th´uccˆa´p r + 1 kh´ac . . . . . 0 th`ıt´ınhtiˆe´p c´acd¯i.nh th´uccˆa´p r +2,c´u nhu thˆe´ d¯ ˆe´nd¯i.nh th´uccˆa´pl´on nhˆa´t V´ıdu T`ım ha.ng cu’a ma trˆa.n 1235 A = 3249 1014 12 Ta c´od¯inh th´u.c con cˆa´p2: = −4 =6 0, v`ac´acd¯inh th´u.ccˆa´p3: . 32 . 123 125 135 235 324 =0; 329 =0; 349 =0; 249 =0 101 104 114 014 suy ra r(A)=2
- 8 . . . + Phuo ng ph´apd`ungph´epbiˆe´nd¯ˆo’iso cˆa´p: biˆe´nd¯ˆo’i ma trˆa.nvˆ`e da.ng bˆa.c thang b1,1 b1,2 b1,r b1,n 0 b2,2 b2,r b2,n B = 00 br,r br,n 00 0 0 00 0 0 . v´oi bi,j =0, ∀i>jhay i>rv`a bii =06 ,i= 1,r th`ı r(A)=r(B)=r. V´ıdu T`ım ha.ng ma trˆa.n 13205 269712 A = − − 2 524 5 148420 1320 5 1320 5 h2−2h1;h3+2h1;h4−h1 0057 2 h4−h3;h2↔h3 016415 A −→ −→ 016415 0057 2 016415 0000 0 suy ra r(A)=3 . + Ngo`aira, c´othˆe’ t`ım ma trˆa.n nghi.ch d¯a’o qua c´acph´epbiˆe´nd¯ˆo’iso cˆa´p: . . . . ’ lˆa.p ma trˆa.n khˆo´i A|E (E c`ung c˜o v´oi A, thu. chiˆe.n c´acph´epbiˆe´nd¯ˆo’isocˆa´p CHI . . . TRENˆ HANG,` nˆe´ud¯uad¯uo. cvˆ`e da.ng E|B th`ı B l`anghi.ch d¯a’ocu’a A. 1 −11| 100 1 −11| 100 h2−2h1,h3−h1 V´ıdu A|E = 211| 010 −→ 03−1 |−210 112| 001 02 1|−101 1 −30| 20−1 1 1 −30| 20−1 h1−h3,h2+h3 h2( ) −→ 050|−31 1 −→5 010|−3/51/51/5 021|−10 1 021|−101 100| 1/53/5 −2/5 h1+3h2,h3−2h2 . . . −→ 010|−3/51/51/5 thu d¯uo. ckˆe´t qua’ nhu c˜u. 011| 1/5 −2/53/5 ` ˆ BAI TA. P . . 1.1. Khˆongt´ınh,ch´ung minh c´acd¯i.nh th´uc sau chia hˆe´t cho 17: 204 323 527 ; 20 9 1 255 55 2 5 . . . 1.2. Ch´ung minh c´acd¯˘a’ ng th´uc sau d¯ˆay(khˆongt´ınhd¯i.nh th´ucb˘a`ng d¯i.nh ngh˜ıa):
- 9 0 xyz 01 1 1 x 0 zy 10z2 y2 . 6 a. = 2 2 v´oi xyz =0 yz0 x 1 z 0 x 2 2 xyz0 1 y x 0 1 xyz b. 1 yzx =(x − y)(y − z)(z − x) 1 zxy 111 c. xyz =(x + y + z)(x − y)(y − z)(z − x) x3 y3 z3 1.3. T`ım x sao cho: 33− x −x xx+1 x +2 a. 273 =0 b. x +3 x +4 x +5 =0 x +1 3x − 7 x x +6 x +7 x +8 1 xx2 x 12 c. 31 x 0 45 1 x +1 2 −4 x 1111 0110 1 xxx 1 x 111 0011 1 a 00 1.4. T´ınh c´acd¯inh th´u.c sau: ; ; 11x 11; . 1001 10b 0 111x 1 1100 100c 1111x 2 3 2 1 xx x a + xx x x +1 xy xz 3 2 2 x x x 1 ab+ xx; xy y +1 yz ; 2 3 ; 2 12x 3x 4x xxc+ x xz yz z +1 3 2 4x 3x 2x 1 axx−x −x 0 xyz 2 x 1 x x 0 y 0 x 2aa 00 x 0 zy 1 x 2 x 0 z 0 t ; ; ; xa2a 00; yz0 x 21xx y 0 z 0 −x 002aa xyz0 xx21 0 t 0 x −x 00 a 2a 12 3 n xaa a 21 2 n− 1 axa a 32 1 n− 2 ; aax a; nn− 1 n − 2 1 aaaax 011 11 cos(x1 − y1) cos(x1 − y2) cos(x1 − yn) 10x x x cos(x2 − y1) cos(x2 − y2) cos(x2 − yn) 1 x 0 x x ; ; cos(xn − y1) cos(xn − y2) cos(xn − yn) 1 xx 0 x 1 xx x0
- 10 a1 −a2 0 00 1+x1y1 1+x1y2 1+x1yn 0 a2 −a3 00 1+x2y1 1+x2y2 1+x2yn 00a3 00 ; 1+xny1 1+xny2 1+xnyn 00 0 an−1 −an 11 1 11+an 212 1 −2 1.5. Cho A = 301 v`a B = 46.T`ım A2,AB,A−1. 012 5 −3 n n n 2 −1 a 1 cos x − sin x 1.6. T`ım c´acma trˆan ; ; . 3 −2 0 a sin x cos x 12 1.7. Cho A = .T`ım f(A)v´o.i f(x)=x2 − 4x +3,f(x)=x2 − 2x +1. 21 1.8. 211 12−2 a. Cho A = 312v`a B = 23 1 . 1 −10 12 2 1. T`ım A−1,B−1. 2. T`ım f(A),f(B)v´o.i f(x)=x2 − x − 1 − 2100 13 57 3200 01 2 −3 b. T`ım ma trˆa.n nghi.ch d¯a’ocu’a A = ; B = . 1134 00 1 2 2 −123 00 0 1 1.9. . . a. T`ım ma trˆa.n vuˆongcˆa´p hai c´ob`ınhphuong b˘a`ng ma trˆa.n khˆong. . . . b. T`ım ma trˆa.n vuˆongcˆa´p hai c´ob`ınhphuong b˘a`ng ma trˆa.nd¯onvi 1.10. T`ım ma trˆa.n X sao cho: 12 35 3 −2 −12 × X = ; X × = ; 34 59 5 −4 56 12−3 1 −30 11−1 1 −13 32−4 ×X = 10 2 7 ; X× 21 0 = 432; 2 −10 10 7 8 1 −11 1 −25 21 −32 −24 × X × = ; 32 5 −3 3 −1 41 21 50 × X × = ; 3 −1 53 61 111 21−1 105 X × 011 − 2 = ; 30 6 −1 −21 001 122 35 15 254 × X + 76 =3 −12; 245 21 −20
- 11 111 1 123 n 011 1 012 n− 1 001 1 × X = 001 n− 2 . 000 1 000 1 1.11. T`ım ha.ng cu’a ma trˆa.n sau: 2111 2 1112 13205 1311 104−1 269712 ; 1141; ; 11 4 56 −5 −2 −524 5 1115 2 −15−6 148420 1111 12314 31−311 32111 135791 2 −17−32 11 1 6; 1 −23−452; 13−253 23−15 2 11 12 25 22 4 3 −27−53 11 0 3 1.12. Biˆe.n luˆa.n theo a sˆo´ ha.ng cu’a c´acma trˆa.n sau: −12 1 1 a −12 a 114 2 a −2 ; 2 −1 a 5 ; 1 a 13; 3 −6(a + 3)(a +7) 110−61 12a 14 − 31 1 4 1436 12 132 a 4101 −1011 2 −1 a2 04 ; ; − 17173 21 10 31 227 22 4 3 02a 4 12 a 11 1.13. T`ım c´acgi´atri. cu’a m d¯ ˆe ’: 34 5 7 1 . 26−34 2 a. r(A)=2v´oi A = 4 2 13 10 0 5 0 21 13 m − 123 11 . 321−11 b. r(A)=3v´oi A = 231 1 1 55202m +1 1436 −10 1 1 c. r(A)=3v´o.i A = − 21 10 02m 4 3114 . m 4101 d. r(A)=2v´oi A = 17173 2243 m 111 . 11m 1 e. r(A)=2v´oi A = 111m 1 m 11
- 12 -ooOoo-
- 13 . . ˆ . . ` ˆ´ ´ Chuong 2. HE. PHUONG TRINH TUYENTINH (2+2) I. C´acd¯i.nh ngh˜ıa . . . . *Tago.i hˆe. phuo ng tr`ınh tuyˆe´nt´ınh m phuo ng tr`ınh n ˆa ’n l`ahˆe. c´oda.ng a1,1x1 + a1,2x2 + ···+ a1,nxn = b1 (1) a2,1x1 + a2,2x2 + ···+ a2,nxn = b2 am,1x1 + am,2x2 + ···+ am,nxn = bm . . trong d¯´o ai,j ,bi (i = 1,m,j = 1,n) l`ac´achˆe. sˆo´ (thu. c ho˘a.cph´uc), x1,x2, ,xn l`ac´ac . . . . . . ˆa’nsˆo´.Hˆe. phuong tr`ınhtuyˆe´n t´ınhd¯uo. cgo.il`ac´onghiˆe.m (hay tuo ng th´ıch)nˆe´u tˆa.p nghiˆe.mcu’a n´okh´acrˆo˜ng. . . . . +Hˆe. (1) c´othˆe’ d¯ u o. cviˆe´tdu´oida.ng ma trˆa.n AX = B trong d¯´o: a a a 1,1 1,2 1,n x1 b1 a a a+2,n A = 2,1 2,2 ; X = x2 ; B = b2 hay . . . x . b am,1 am,2 am,n n m a1,1 a1,2 a1,n b1 . . . a2,1 a2,2 a+2,n b2 du´oida.ng ma trˆa.nmo’ rˆo.ng: A = , khi d¯´oha.ng am,1 am,2 am,n bm . . . . r(A)cu’a A d¯ u o. cgo.il`aha.ng cu’ahˆe. phuo ng tr`ınh (1) II. Hˆe. Cramer . . . . . *Hˆe. (1) c´osˆo´ phuong tr`ınh b˘a`ng sˆo´ nghiˆe.m(m = n) v`ad¯i.nh th´uc det(A)=0d¯uo. c go.il`ahˆe. Cramer. D +Hˆe Cramer c´onghiˆem duy nhˆa´td¯u.o.c x´acd¯inh nhu. sau: ∀i = 1,n,x = i , trong . . . . i D . . . . . d¯ ´o D = det(A), c`on Di l`ad¯i.nh th´uc thu d¯uo. ct`uD b˘a`ng c´ach thay cˆo.tth´ui b˘a`ng . cˆo.thˆe. sˆo´ tu. do. x1 +2x2 +3x3 =6 V´ıdu Gia’ihˆe.: 2x1 − x2 + x3 =2 3x1 + x2 − 2x3 =2 12 3 Do D = 2 −11 =30=0,hˆe6 c´onghiˆem duy nhˆa´t(1, 1, 1): . . 31−2 62 3 16 3 126 1 1 1 x = 2 −11 =1;y = 22 1 =1;z = 2 −12 =1 30 30 30 21−2 32−2 312 III. C´acd¯i.nh l´yvˆe` nghiˆe.mcu’ahˆe. (Kronecker-Kapeli) . . + (1) c´onghiˆe.m (tuong th´ıch) khi v`achı’ khi r(A)=r(A). + (1) c´onghiˆe.m duy nhˆa´t (x´acd¯i.nh) khi v`achı’ khi r(A)=r(A)=n. +nˆe´u r(A)=r(A)=r<nth`ı(1) c´ovˆosˆo´ nghiˆe.m v`ac´acth`anhphˆa`n nhiˆe.m phu. thuˆo.c n − r tham sˆo´ tu`y´y.
- 14 ax1 + x2 + x3 =1 V´ıdu Biˆe.n luˆa.n theo a sˆo´ nghiˆe.mcu’ahˆe.: x1 + ax2 + x3 =1 x1 + x2 + ax3 =1 . D`ung c´acph´ep biˆe´nd¯ˆo’isocˆa´pd¯ˆe’ x´acd¯i.nh ha.ng cu’a A v`a A a 11| 1 11a | 1 ↔ A = 1 a 1 | 1 h−→1 h3 1 a 1 | 1 11a | 1 a 11| 1 11 a | 1 11 a | 1 − h−→2 h1 0 a − 11− a | 0 h−→3+h2 0 a − 11− a | 0 − h3 ah1 01− a 1 − a2 | 1 − a 002− a − a2 | 1 − a 2 . . . +Nˆe´u2− a − a = 0, c´o2 tru`o ng ho. p: 111| 1 a = 1 th`ı: A −→ 000| 0 ⇒ r(A)=r(A)=1< 3, hˆe. c´ovˆosˆo´ 000| 0 nghiˆe.m phu. thuˆo.c 2 tham sˆo´ tu`y´y. 11−2 | 1 a = −2 th`ı: A −→ 0 −33| 0 ⇒ r(A)=2<r(A =3,hˆe. vˆo 00 0| 3 nghiˆe.m. +Nˆe´u2− a − a2 =06 ⇔ a =16 ,a=6 −2, th`ı r(A)=r(A)=3,hˆe. c´onghiˆe.m duy nhˆa´t. . . IV. Phuong ph´apgia’ihˆe. . . . . . . . . . + C´acph´epbiˆe´nd¯ˆo’isocˆa´p cho hˆe. tuong d¯uong (tuong ´ung v´oi c´acph´epbiˆe´nd¯ˆo’i . theo h`angcu’a ma trˆa.nmo’ rˆo.ng): . . − D- ˆo’ichˆo˜ hai phuong tr`ınhcho nhau (d¯ˆo’ichˆo˜ hai h`angcu’a ma trˆa.n) . . . . − Nhˆanhai vˆe´ cu’aphuong tr`ınh n`aod¯´ov´oimˆo.tsˆo´ kh´ac0 (nhˆanc´acphˆa`ntu’ . trˆen mˆo.t h`angcu’a ma trˆa.nv´oimˆo.tsˆo´ kh´ac0) . . . . . . . − Cˆo.ng t`ung vˆe´ cu’amˆo.tphuong tr`ınhv´oimˆo.tphuong tr`ınhkh´acnhˆanv´oi . mˆo.tsˆo´ (cˆo.ng mˆo.t h`angv´oibˆo.isˆo´ mˆo.t h`angkh´ac) ´ 1. Ap du.ng d¯i.nh l´y Carmer . . Nˆe´uhˆe. phuong tr`ınhtuyˆe´n t´ınhl`ahˆe. Cramer, c´othˆe’ ´apdu. ng d¯i.nh l´y Carmer − − ho˘a.c t`ımma trˆa.n A 1, suy ra X = A 1B. 2x +3y +2z =9 . . V´ıdu Gia’ib˘a`ng phuong ph´apma trˆa.n nghi.ch d¯a’o: x +2y − 3z =14 3x +4y − z =16 23 2 Do det(A)= 12−3 = −6 =0nˆe6 ` hˆe l`aCramer. . 34 1 A A A 14 5 −13 1 1,1 2,1 3,1 1 V´o.i A−1 = A A A = −10 −48 det(A) 1,2 2,2 3,2 −6 A1,3 A2,3 A3,3 −21 1 14 5 −13 9 2 x =2 1 nˆen X = A−1B = − −10 −48 14 = 3 , suy ra y =3 6 −21 1 16 −2 z = −2.
- 15 2. Phu.o.ng ph´apGauss (khu’. dˆa` nˆa’nsˆo´) . . D`ung c´acph´ep biˆe´nd¯ˆo’isocˆa´p theo c´ach`ang,biˆe´nd¯ˆo’i ma trˆa.nmo’ rˆo.ng A th`anh . . ma trˆa.n A1 c´onhiˆe` u phˆa`ntu’ 0 (nhu ma trˆa.nbˆa.c thang), khi d¯´o r(A)=r(A1)v`a r(A)=r(A1). +nˆe´u r(A1) <r(A1), th`ıhˆe. vˆonghiˆe.m . . . . . . . +nˆe´u r(A1)=r(A1)=r th`ılˆa.phˆe. phuong tr`ınh m´oi (tuong d¯uong hˆe. d¯˜acho) sau . . . kho bo’ c´ach`angm`amo.i phˆa`ntu’ d¯` ˆe ub˘a`ng 0. Gia’ihˆe. n`ay(r phuong tr`ınh, n ˆa ’n . . sˆo´)b˘a`ng c´ach cho.n r ˆa ’ncoba’nv`an − r ˆa ’n khˆongco ba’n (thay b˘a`ng tham sˆo´ tu`y y),´ nˆe´u r = n th`ıhˆe. c´onghiˆe.m duy nhˆa´t. . . V´ıdu Gia’i c´achˆe. phuong tr`ınh sau: x1 − 3x2 +2x3 = −1 x1 +9x2 +6x3 =3 x1 +3x2 +5x3 =1 1 −32−1 1 −321 1 −32−1 − h ×1/2 A = 1963 h−→2 h1 01244 2−→ 0311, − − 1351 h3 h1 0632 h3 h2 0010 x =0 x1 − 3x2 +2x3 = −1 x1 = −1+3x2 − 2x3 1 1 suy ra 3x2 + x3 =1 ⇒ 3x2 =1− x3 ⇒ x = 2 3 x =0 x =0 3 3 x3 =0 x1 − 3x2 +2x3 − x4 =2 2x1 +7x2 − x3 = −1 4x1 + x2 +3x3 − 2x4 =1 1 −32−12 1 −32−12 − − B = 27−10−1 h2−→2h1 013−52−5 h−→3 h2 − 41 3−21 h3 4h1 013−52−7 1 −32−12 013−52−5 = B1.Dor(B)=r(B1)=2< 3=r(B1)=r(B), hˆe. 00 0 0−2 vˆonghiˆe.m. x1 +5x2 +4x3 +3x4 =1 2x − x +2x − x =0 1 2 3 4 5x1 +3x2 +8x3 + x4 =1 4x1 +9x2 +10x3 +5x4 =2 15 4 31 15431 2 −12−10 h3−h1−2h2 2 −12−10 C = −→ 53 8 11 h4−2h1−h2 00000 491052 00000 h −2h 15 4 3 1 2−→ 1 ,t´u.c l`a: bo’ h3,h4 0 −11 −6 −7 −2
- 16 x1 +5x2 +4x3 +3x4 =1 −11x2 − 6x3 − 7x4 = −2. 14 2 1 x1 = − α + β + 11 11 11 6 7 2 Chon x = α, x = β, ta suy ra: x = − α − β + . 3 4 2 11 11 11 x3 = α x4 = β ax + y + z =1 V´ıdu. 2. Gia’i v`abiˆe.n luˆa.n theo a: x + ay + z = a 2 x + y + az = a a 11 1 11aa2 11 aa2 ↔ − A = 1 a 1 a h−→3 h1 1 a 1 a h−→2 h1 0 a − 11− aa− a2 − 11aa2 a 11 1 h3 ah1 01− a 1 − a2 1 − a3 11 aa2 h−→3+h2 0 a − 11− aa− a2 , suy ra: 002− a − a2 1+a − a2 − a3 *Nˆe´u2− a − a2 =0⇔ (a =1)∨ (a = −2) . . . . . +Nˆe´u a = 1, th`ı A → (1 1 1 1), tuong d¯uong v´oi x + y + z = 1 nˆenc´ovˆosˆo´ nghiˆe.m . da.ng (1 − α − β; 1; 1) v´oi α, β tu`y´y. 11−24 +Nˆe´u a = −2, th`ı A → 0 −33−6 suy ra r(A)=2< 3=r(A)nˆen hˆe. vˆo 00 0 3, nghiˆe.m. *Nˆe´u2− a − a2 =06 ⇔ (a =1)6 ∧ (a =6 −2) 2 11 aa h2:a−1 01−1 −a . . . . A −→ 2 , nˆen hˆe. d¯˜acho tuong ´ung v´oi: − − 2 (a +1) h3:2 a a 00 1 a +2 a +1 2 x1 = − x + y + az = a a − 2 y − z = −a 1 ⇔ x2 = (a +1)2 a +2 z = (a +1)2 a − 2 x = 3 a +2 ax + y + z =1 V´ıdu. 3. Gia’i v`abiˆe.n luˆa.n theo a, b: x + by + z =3 x +2by + z =4 a 11 411 D = det(A)= 1 b 1 =(1− a)b; Dx = 3 b 1 = −2b +1; 12b 1 42b 1 a 41 a 14 Dy = 131 =1− a; Dz = 1 b 3 =4b − 2ab − 1 141 12b 4
- 17 a =16 +Nˆe´u D =(1− a)b =06 ⇔ ,hˆe. l`aCramer, c´onghiˆe.m duy nhˆa´t: b =06 −2b +1 x1 = (1 − a)b 1 x = 2 b 4b − 2ab − 1 x3 = (1 − a)b x + y + z =1 x + y + z =4 . +Nˆe´u a =1,hˆe. tro’ th`anh: x + by + z =3 ⇔ (b − 1)y = −1 , th`ı:: x +2by + z =4 (2b − 1)y =0 x =2− α 1 x + y + z =0 − Nˆe´u2b − 1=0⇔ b = : ⇔ y =2 , α tu`y´y. 2 y =2 z = α x + y + z =4 1 − Nˆe´u2b−1 =06 ⇔ b =6 : (b − 1)y = −1 vˆonghiˆemv`ı(b−1)0 = −1 2 . y =0 ax − y + z =4 . +Nˆe´u b =0,hˆe. tro’ th`anh: x + z =3 vˆonghiˆe.m x + z =4 . . V. Hˆe. phuong tr`ınhtuyˆe´n t´ınhthuˆa` n nhˆa´t . . * Hˆe. phuo ng tr`ınh tuyˆe´nt´ınh thuˆa` n nhˆa´t l`ahˆe. c´oda.ng AX = 0 (II) (B l`ama trˆa.n to`ansˆo´ 0), khi d¯´o r(A)=r(A), hˆe. luˆonluˆonc´onghiˆe.m: . . +nˆe´u r(A)=n,hˆe. c´onghiˆe.m duy nhˆa´t nghiˆe.mtˆa` m thu`o ng x1 = x2 = ··· = xn =0; +nˆe´u r(A) <n,hˆe. c´ovˆosˆo´ nghiˆe.m, c´acth`anhphˆa`ncu’a nghiˆe.m phu. thuˆo.c n − r(A) . . tham sˆo´, nˆenc´onghiˆe.m kh´acnghiˆe.m khˆong(nghiˆe.m khˆongtˆa` m thu`o ng). . . . . . +V´oihˆe. c´o n phuong tr`ınh, n ˆa ’nsˆo´,hˆe. c´onghiˆe.m khˆongtˆa`mthuong khi v`achı’ khi . . det(A) =6 0 v`ac´onghiˆe.m duy nhˆa´ttˆa`mthuong khi v`achı’ khi det(A)=0. ax1 + x2 + ···+ xn−1 + xn =0 x1 + ax2 + ···+ xn−1 + xn =0 V´ıdu T`ım a d¯ ˆe ’ hˆe. =0 c´onghiˆe.m khˆongtˆa`m x1 + x2 + ···+ axn−1 + xn =0 x1 + x2 + ···+ xn−1 + axn =0. thu.o.ng
- 18 a 1 11 1 a 11 det(A)= 11 a 1 11 1 a a + n − 1 a + n − 1 a+ n − 1 a + n − 1 1 a 11 h1+P hi = i=1 11 a 1 11 1 a 11 11 1 a 11 =(a + n − 1) 11 a 1 11 1 a 11 11 0 a − 1 00 hi−h1 n−1 = (a + n − 1) =(a + n − 1)(a − 1) i=1 00 a− 10 00 0 a − 1 a =1− n Hˆe c´onghiˆem khˆongtˆa`mthu.`o.ng khi det(A)=0⇔ . . a =1. +nˆe´u (α1; α2; ; αn−1; αn)v`a(β1; β2; ; βn−1; βn) l`anghiˆe.mcu’ahˆe. (II) th`ı ∀h, k ∈ R :(hα1 + kβ1; hα2 + kβ2; ; hαn−1 + kβn−1; hαn + kβn) c˜ung l`anghiˆe.mhˆe. (II). . . . . . . +Tru`ong ho.p r(A) <n(sˆo´ ˆa’ncu’ahˆe.)th`ır(A)ˆa’ncoba’nd¯uo. cbiˆe’udiˆ˜en qua . . n − r(A)ˆa’n khˆongco ba’n (lˆa´y gi´atri. tu`y´y).Nˆe´ucho.n n − r(A)ˆa’n khˆongco ba’n . . . tuong ´ung theo n − r(A) th`anhphˆa`ncu’a n − r(A)bˆo. sˆo´: (1; 0; 0; ; 0); (0; 1; 0; ; 0); (0; 0; 1; ; 0); ; (0; 0; 0; ;1) . . . th`ı n − r(A) nghiˆe.mcu. thˆe’ cu’ahˆe. (II) d¯uo. cgo.il`amˆo.thˆe. nghiˆe.mco ba’ncu’a hˆe x1 +2x2 − 2x3 + x4 =0 . 2x1 +4x2 +2x3 − x4 =0 V´ıdu T`ım hˆe. nghiˆe.mcoba’ncu’a x1 +2x2 +4x3 − 2x4 =0 4x1 +8x2 − 2x3 + x4 =0.
- 19 − − 12 21 12 21 24 2 −1 h2−2h1 00 6 −3 h3−h2 12−21 . A = −→ −→ u´ng 12 4 −2 h3−h1 00 6 −3 h2:2 00 2 −1 h −h 48−21 h4−4h1 00 6 −3 4 2 . v´oihˆe.: x +2x − 2x + x =0 x = −2x 1 2 3 4 ⇔ 1 2 2x3 − x4 =0 x4 =2x3. + Cho.n(x2,x3)=(1, 0), ta c´o:nghiˆe.m(−2; 1; 0; 0) + Cho.n(x2,x3)=(0, 1), ta c´o:nghiˆe.m (0; 0; 1; 2) . . . * Gia’ith´ıch c´acht`ım ma trˆa.nghi.ch d¯a’oo’ phˆa` n IV, chuo ng 1 a1,1 a1,2 a1,3 a1,n a2,1 a2,2 a2,3 a2,n Cho ma trˆa.n vuˆong A = c´odet(A) =0.X´6 et hˆe. an,1 an,2 an,3 an,n n phu.o.ng tr`ınh 2n ˆa ’n: a1,1x1 + a1,2x2 + a1,3x3 + ···+ a1,nxn + xn+1 =0 a2,1x1 + a2,2x2 + a2,3x3 + ···+ a2,nxn + xn+2 =0 a3,1x1 + a3,2x2 + a3,3x3 + ···+ a3,nxn + xn+3 =0 an,1x1 + an,2x2 + an,3x3 + ···+ an,nxn + xn+1 =0 c´oda.ng ma trˆa.n A × X + X0 =0⇔ A × X = −X0 (1) x1 xn+1 x2 xn+2 v´o.i X = x3 v`a X0 = xn+3 . . . . . . xn x2n v`ı det(A) =06 , ∃A−1 nˆen:(1)⇔ X = −A−1 × X0 ⇔ X + A−1 × X0 = 0 (*) | a1,1 a1,2 a1,3 a1,n 100 0 a2,1 a2,2 a2,3 a2,n | 010 0 Hˆe. c´oma trˆa.nhˆe. sˆo´: a3,1 a3,2 a3,3 a3,n | 001 0 =(A|E) an,1 an,2 an,3 an,n | 000 1 . . . . . Gia’ su’ qua c´acph´ep biˆe´nd¯ˆo’isocˆa´p trˆenc´ach`ang,ta d¯uad¯uo. c ma trˆa.nvˆe` da.ng 100 0 | b1,1 b1,2 b1,3 b1,n 010 0 | b2,1 b2,2 b2,3 b2,n 001 0 | b3,1 b3,2 b3,3 b3,n =(E|B) 000 1 | bn,1 bn,2 bn,3 bn,n
- 20 . . u´ng v´oihˆe.: x1 + b1,1xn+1 + b1,2xn+2 + b1,3xn+3 + ···+ b1,nx2n =0 x2 + b2,1xn+1 + b2,2xn+2 + b2,3xn+3 + ···+ b2,nx2n =0 x3 + b3,1xn+1 + b3,2xn+2 + b3,3xn+3 + ···+ b3,nx2n =0 xn + bn,1xn+1 + bn,2xn+2 + bn,3xn+3 + ···+ bn,nx2n =0 0 − c´oda.ng X + B × X = 0, suy ra B = A 1 BAI` TAˆ. P 3x − 5y +2z +4t =2 2x + y − z =1 . . 2.1. Gia’ic´achˆe. phuong tr`ınhsau: 7x − 4y + z +3t =5 x − y + z =2 5x +7y − 4z − 6t =3 4x +3y + z =3 x + y − 3z = −1 2x +3y − z +5t =0 x − 2y +3z − 4t =4 2x + y − 2z =1 3x − y +2z − 7t =0 y − z + t = −3 x +2y − 3z =1 4x + y − 3z +6t =0 x +3y − 3t =1 x + y + z =3 x − 2y +4z − 7t =0 −7y +3z +3t = −3 x − y +2z − 3t =1 2x + y − 3z =4 x +3y +4z =8 x +4y − z − 2t = −2 x +2y + z =1 2x + y − z =2 x − 4y +3z − 2t = −2 3x − 3y +2z =11 2x +6y − 5z =4 x − 8y +5z − 2t = −2 2x +3y − z + t =2 3x +4y +5z +7t =1 x + y +5z = −7 2x +3y + z =4 2x +6y − 3z +4t =2 x +3y + z =5 2x +3y +2z =3 4x +2y +13z +10t =0 2x + y + z =2 2x +3y =5 2x +21z +13t =3 2x +3y − 3z =14 2x − 5y +4z +3t =0 3x + y − 3z + t =1 x +2y +3z − t =1 3x − 4y +7z +5t =0 2x − y +7z − 3t =2 3x +2y + z − t =1 4x − 9y +8z +5t =0 x +3y − 2z +5t =3 2x +3y + z + t =1 3x − 2y +5z − 3t =0 3x − 2y +7z − 5t =3 5x +5y +5z =2 8x +6y +5z +2t =21 x1 + x2 =1 3x +3y +2z + t =10 x1 + x2 + x3 =4 4x +2y +3z+=8 x2 + x3 + x4 = −3 3x +5y + z + t =15 x3 + x4 + x5 =2 7x +4y +5z +2t =18 x4 + x5 = −1 x1 +2x2 +3x3 +4x4 =0 7x +14x +20x +27x =0 1 2 3 4 5x1 +10x2 +16x3 +19x4 = −2 3x1 +5x2 +6x3 +13x4 =5 2.2. Gia’iv`abiˆe.n luˆa.n theo a c´achˆe. sau:
- 21 (a +1)x + y + z =1 ax + y + z + t =1 x +(a +1)y + z = a x + ay + z + t = a 2 2 x + y +(a +1)z = a x + y + az + t = a x − y + az + t = a − − 2 11 1 x 1 x + ay − z + t = −1 2 −10−3 y 2 × = ax + ay − z − t = −1 30−11 z −3 22−2 a t −6 x + y + z + t = −a ax1 − 3x2 + x3 = −2 . . 2.3. Cho hˆe. phuong tr`ınh ax1 + x2 +2x3 =3 3x1 +2x2 + x3 = b. . . . a. T`ım a d¯ ˆe ’ hˆe. trˆen l`ahˆe. Cramer; ´ung v´oi gi´atri. cu’a a v`ua t`ım,t`ımnghiˆe.mcu’ahˆe. theo b. b. T`ım a, b d¯ ˆe ’ hˆe. trˆen vˆonghiˆe.m. c. T`ım a, b d¯ ˆe ’ hˆe. trˆen c´ovˆosˆo´ nghiˆe.m, t`ımnghiˆe.mtˆo’ng qu´atcu’ahˆe . . 2.4. T`ım m d¯ ˆe ’ c´achˆe. phuong tr`ınhsau d¯ˆay: a. c´onghiˆe.m 23 1 x 7 36 −9 x 37−6 × y = −2 ; 48 × = 12 ; y 58 1 x m 27 m 32 5 x 1 375 x −m 24 6 × y = 3 ; 231 × y = 2 ; 57m z 5 693 z 5 mx +2y +3z +2t =3 3x +4y +5z +7t =1 2x + my +3z +2t =3 2x +6y − 3z +4t =2 ; 2x +3y + mz +2t =3 4x +2y +13z +10t = m 2x +3y +2z + mt =3 5x +21z +13t =3 2x +3y +2z +3t = m 2x − y + z − t =1 x + y +(1− m)z = m +2 2x − y − 3t =2 b. vˆonghiˆe.m: ; (1 + m)x − y +2z =0 3x − z + t = −3 2x − my +3z = m +2 2x +2y − 2z + mt = −6 mx1 + x2 + x3 + ···+ xn =0 x1 + mx2 + x +3+···+ xn =0 c. vˆod¯i.nh: x1 + x2 + mx3 + ···+ xn =0 x1 + x2 + x3 + ···+ mxn =0 3x +2y + z + t =1 3x +2y + z =3 x + my − z +2t =0 2x +3y + z + t =1 ; mx + y +2z =3 ; 2x − y + mz +5t =0 x +2y +3z − t =1 mx − 3y + z = −2 x +10y − 6z + t =0 5x +5y +2z =2m +1
- 22 d. c´onghiˆe.m duy nhˆa´t: x +4y +3z +6t =0 x +3y − z + t =1 x + y + z + mt =1 −x + z + t =0 3x +3y − z + mt =2 x + my + z + t =1 ; ; 2x + y − z =0 2x +2y + z + t =3 mx + y + z + t =1 2y + mx =0 5x +3y +2t =1 x + y + mz + t =1 2x +5y +3z +7t =0 . 2.5. Ch´ung minh hˆe. sau c´onghiˆe.m duy nhˆa´t, t`ımnghiˆe.m d¯´o: x2 + x3 + x4 + ···+ xn−1 + xn =1 x1 + x3 + x4 + ···+ xn−1 + xn =2 x1 + x2 + x4 + ···+ xn−1 + xn =3 x1 + x2 + x3 + x4 + ···+ xn−1 = n 2.6. T`ım d¯iˆ`eukiˆe.n theo a d¯ ˆe ’ hˆe. sau c´onghiˆe.m duy nhˆa´t x1 + ax2 =0 x1 +(1+a)x2 + ax3 =0 x2 +(1+a)x3 + ax4 =0 x3 +(1+a)x4 + ax5 =0 x4 +(1+a)x5 =0 . . 2.7. Biˆe.n luˆa.n theo a sˆo´ nghiˆe.mcu’ahˆe. phuong tr`ınh: (a − 3)x + y + z =0 ax + ay + z = a x +(a − 3)y + z =0; ax + y + az =1; x + y +(a − 3)z =0 x + ay + az =1 x − y + az + t = a ax + ay +(a +1)z = a x + ay − z + t = −1 ax + ay +(a − 1)z = a ; ax + ay − z − t = −1 (a +1)x + ay +(2a +3)z =1 x + y + z + t = −a . . . . 2.8. T`ım nghiˆe.m nguyˆenduong (nˆe´u c´o)cu’ahˆe. phuong tr`ınhsau: x + y + z = 100 x +2y +3z =14 ; ; x +15y +25z = 500 2x +3y − z =5 x − y + z + t =2 x +3y − 3z =1 ; 2x + y − 3z +2t =2 3x − 3y +4z =4 3x − 2y + z + t =3 . 2.9. T`ım c´acd¯ath´ucbˆa.c3f(x)biˆe´t: a. f(−1) = 0; f(1) = 4; f(2) = 3; f(3) = 16; b. f(−1) = 5; f(1) = 5; f(3) = 45; f(−4) = −25. . . . 2.10. T`ım nghiˆe.mtˆo’ng qu´atv`ahˆe. nghiˆe.mcoba’ncu’ahˆe. phuong tr`ınhsau:
- 23 x + y − 4z =0 2x − y +5z +7t =0 2x +9y +6z =0 4x − 2y +7z +5t =0; ; 3x +5y +2z =0 2x − y + z − 5t =0 4x +7y +5z =0 x +2y +4z − 3t =0 x +8z +7t =0 3x +5y +6z − 4t =0 2x + y +4z + t =0 ; 4x +5y − 2z +3t =0 3x +2y − z − 6t =0 3x +8y +24z − 19t =0 7x +4y +6z − 5t =0 2.11. . . a. Trong mˆo.t x´ınghiˆe.psa’n xuˆa´t, c´o15 cˆongnhˆand¯uo. c chia l`am3 bˆa.c (I,II,III), . . . . . . huo’ ng luong th´anglˆa`nluo. t l`a: 600.000, 500.000, 400.000 d¯ˆo`ng. Mˆo˜i th´angx´ı . . nghiˆe.p ph´at7,7 triˆe.ud¯ˆo`ng tiˆe` nluong. Ho’i trong x´ı nghiˆe.pˆa´y, sˆo´ cˆongcˆongmˆo˜i bˆa.cc´othˆe’ l`abao nhiˆeu? . b. Mˆo.tho. p t´acx˜anˆongnghiˆe.p c´o300 ha d¯ˆa´t, 850 cˆonglao d¯ˆo.ng v`a65 triˆe.ud¯ˆo`ng . . tiˆe` nvˆo´n d`anhcho sa’n xuˆa´tvu. h`ethu v´oidu. d¯ i .nh trˆo`ng c´acloa.i cˆayI,II,III c´ochi ph´ısa’n xuˆa´t cho mˆo˜i ha giao trˆo`ng nhu. sau: Loa.i cˆay Vˆo´nb˘a`ng tiˆe` n (d¯ˆo`ng) Lao d¯ˆo.ng (cˆong) I 200.000 2 II 150.000 3 III 400.000 5 -ooOoo-
- 24 Chu.o.ng 3 HAM` NHI`EˆUBIEˆ´N&T´ICH PHANˆ KEP´ I. H`amnhiˆe` ubiˆe´n 1. Kh´ainiˆe.m * Cho D ⊂ R2.Mˆo.t ´anhxa. f : D → R (x, y) 7→ f(x, y)=z ∈ R . . . . d¯ u o. cgo.il`ah`amhai biˆe´n x´acd¯i.nh trˆen D, D d¯ u o. cgo.il`amiˆ`en x´acd¯i.nh cu’a h`amhai biˆe´n f(x, y). V´ıdu +Miˆe` n x´acd¯i.nh cu’a h`am z = f(x, y)=p1 − x2 − y2 l`atˆa.p D = (x, y) ∈ R2 : x2 + y2 ≤ 1 (h`ınhtr`ontˆam O b´ank´ınh1). +Miˆe` n x´acd¯i.nh cu’a h`am z = f(x, y)=ln(x+y) l`atˆa.p D = (x, y) ∈ R2 : x + y>0 . . . (nu’ am˘a.t ph˘a’ ng n˘a`m ph´ıatrˆen d¯u`o ng th˘a’ ng y = −x trˆen m˘a.t ph˘a’ ng xOy. . . * Cho h`amhai biˆe´n z = f(x, y). Trˆen m˘a.t ph˘a’ ng Oxy,mˆo˜ic˘a.p(x, y)d¯uo. cbiˆe’udiˆe˜n . bo’ imo.td¯iˆe’m M(x, y), nˆen ta c´othˆe’ xem z = f(x, y) l`ah`amc´acd¯iˆe’m M(x, y), k´y hiˆe` u z = f(M). * Cho h`amhai biˆe´n z = f(x, y) c´omiˆe` n x´acd¯i.nh D. Trong khˆonggian Oxyz,x´et c´acd¯iˆe’m P (x, y, z) tho’a m˜an(x, y) ∈ D v`a z = f(x, y). Khi M cha.y trˆenmiˆe` n D, . . c´acd¯iˆe’m P va.ch trong khˆonggian mˆo.tm˘a.t cong d¯uo. cgo.il`ad¯` ˆo thi. cu’a h`am hai biˆe´n x = f(x, y). n * Cho D ⊂ R = {(x1,x2, ,xn):xi ∈ R,i =1, , n}.Mˆo.t ´anhxa. f : D → R (x1,x2, ,xn) 7→ f(x1,x2, ,xn)=z ∈ R . . . . d¯ u o. cgo.il`ah`am n biˆe´n f(x1,x2, ,xn) x´acd¯i.nh trˆen D (D d¯ u o. cgo.il`amiˆ`en x´acd¯i.nh). . * Cho h`amhai biˆe´n z = f(x, y) x´acd¯i.nh trong khoa’ng ho’ U cu’a Mo(xo,yo) (khˆong . . . cˆa`n x´acd¯i.nh ta.i Mo). Sˆo´ L d¯ u o. cgo.il`agi´o iha.ncu’a f(x, y) khi M(x, y) dˆa` n . d¯ ˆe´n Mo(xo,yo)nˆe´uv´oimo.i d˜ayd¯iˆe’m Mn(xn,yn) thuˆo.c U dˆa`nd¯ˆe´n Mo(xo,yo), ta d¯` ˆe u c´o: lim f(xn,yn) → L.Tak´yhiˆeu: n→∞ . lim f(x, y)=L. x→xo y→yo . . * H`amsˆo´ z = f(x, y) x´acd¯i.nh trong miˆ`en D d¯ u o. cgo.il`aliˆentu.cta.i Mo(xo,yo) ∈ D nˆe´u: lim f(x, y)=f(xo,yo). x→xo y→yo
- 25 2. D- a.o h`amv`avi phˆanh`amnhiˆe` ubiˆe´n 2.1. D- a.o h`amriˆeng . * Cho h`amsˆo´ z = f(x, y) x´acd¯i.nh trˆenkhoa’ng ho’ U cu’a Mo(xo,yo), khi d¯´o∆x = . . . . x−xo v`a∆y = y−yo d¯ u o. cgo.ilˆa`nluo. tl`asˆo´ gia cu’abiˆe´nsˆo´ x v`a y,∆xz = f(xo+ . . . . ∆x, yo)−f(xo,yo)v`a∆yz = f(xo,yo+∆y)d¯uo. cgo.ilˆa`nluo. tl`asˆo´ gia riˆeng cu’a h`am z = f(x, y) theo x v`atheo y ta.i Mo(xo,yo), c`on∆z = f(xo +∆x, yo +∆y)−f(xo ,yo) . . d¯ u o. cgo.il`asˆo´ gia to`anphˆa` ncu’a h`am z = f(x, y) ta.i Mo(xo,yo). ∆xz ∆yz . . . . *Nˆe´u lim v`a lim tˆo`nta.ih˜uuha.n th`ıc´acgi´oiha.n d¯´od¯uo. cgo.i l`ac´ac ∆x→0 ∆x ∆y→0 ∆y d¯ a.o h`amriˆeng cu’a h`am x = f(x, y) ta.i (xo,yo) cu’abiˆe´n x v`abiˆe´n y,k´y . . hiˆe.ulˆa`nluo. t l`a: 0 0 ∂z ∆xz zx(xo,yo)=fx(xo,yo)= (xo,yo) = lim ∂x ∆x→0 ∆x 0 0 ∂z ∆yz zy(xo,yo)=fy(xo,yo)= (xo,yo) = lim ∂y ∆y→0 ∆y *Nˆe´u h`am z = f(x, y) c´oc´acd¯a.o h`amriˆengtheo biˆe´n x v`abiˆe´n y ta.i ∀(x, y) ∈ D, ta n´oi z = f(x, y) c´oc´acd¯a.o h`amriˆengtheo biˆe´n x v`atheo biˆe´n y trong miˆ`en D,k´yhiˆe.u l`a: ∂z ∂z f 0 (x, y)=z0 = ; f 0 (x, y)=z0 = x x ∂x y y ∂y V´ıdu T´ınh c´acd¯a.o h`amriˆeng cu’a + z = xy ,x>0: 0 y 0 y−1 zx =(x )x = yx ; 0 y 0 y zy =(x )x = x . ln x x + z = e y : 0 0 x x x x 1 0 y y y zx = e = e . = e . ; x y x y 0 0 0 x x x x x x x y y y − − y zy = e = e . = e . 2 = 2 .e y y y y y + z = Arctg xy; (xy)0 y z0 = (Arctg xy)0 = x = ; x x 1+(xy)2 1+x2y2 (xy)0 x z0 = y = y 1+(xy)2 1+x2y2 2.2. Vi phˆan 0 0 . * Vi phˆanto`anphˆa`ncu’a h`amhai biˆe´n z = f(x, y) l`a: dz = zxdx + zydy,c´othˆe’ u´ng . . . du.ng d¯ˆe’ t´ınhgˆa`nd¯´ung gi´atri. cu’a h`amsˆo´ ph´ucta.p theo cˆongth´ucsˆo´ gia h˜uuha.n nhu. sau: 0 0 f(xo +∆x, yo +∆x) ' fx(xo,yo) · ∆x + fy (xo,yo) · ∆y + f(xo,yo)
- 26 V´ıdu T´ınh gˆa`n d¯´ungc´acsˆo´sau: a. A =(0.998)3.001; b. B = p(4.001)2 +(2.997)2 y a. X´et z = f(x, y)=x ta.i Mo(1; 3). Ta c´o: + f(x, y)=xy ⇒ f(1, 3) = 3.12 =3 0 y−1 0 2 + fx(x, y)=yx ⇒ fx(1, 3)=3.1 =3 0 y 0 3 + fy(x, y)=x . ln x ⇒ fy (1, 3)=1 . ln 1 = 0 Cho.n∆x = −0.002, ∆y =0.001, khi d¯´o: 3+0.001 A =(1− 0.002) = f(1 − 0.002, 3+0.001) = f(xo +∆x, yo +∆y) 0 0 ' fx(xo,yo) · ∆x + fy (xo,yo) · ∆y + f(xo,yo)=3· (−0.002) + 0 · (0.001) + 1 = 0.994 2 2 b. X´et z = f(x, y)=px + y ta.i M√o(4, 3). Ta c´o: + f(x, y)=px2 + y2 ⇒ f(4, 3) = 42 +32 =5 0 x 0 4 4 + fx(x, y)= ⇒ fx(4, 3) = = =0.8 px2 + y2 42 +32 5 0 y 0 3 3 + fy(x, y)= ⇒ fx(4, 3) = = =0.6 px2 + y2 42 +32 5 Cho.n∆x =0.001, ∆y = −0.003, khi d¯´o: 2 2 2 2 B = p(4.001) +(2.997) = p(4 + 0.0001) +(3− 0.003) = f(xo +∆x, yo +∆y) 0 0 ' fx(xo,yo) · ∆x + fy(xo,yo) · ∆y + f(xo,yo) =0.8 · 0.001 + 0.6 · (−0.003) + 5 = 4.999 0 0 . . *Nˆe´u c´acd¯a.o h`amriˆeng zx,zy (d¯uo. cgo.il`ad¯ a.o h`amriˆeng cˆa´p1)c˜ung c´od¯a.o . . h`amriˆeng th`ıc´acd¯a.o h`amriˆeng d¯´od¯uo. cgo.il`ad¯ a. o h`amriˆengcˆa´p2cu’a . . . z = f(x, y), d¯uo. ck´yhiˆe.u v`ax´acd¯i.nh nhu sau: ∂2f z00 = f 00 (x, y)= =(z0 )0 ; xx xx ∂x2 x x ∂2f z00 = f 00 (x, y)= =(z0 )0 ; xy xy ∂x∂y x y ∂2f z00 = f 00 (x, y)= =(z0 )0 ; yx yx ∂y∂x y x ∂2f z00 = f 00 (x, y)= =(z0 )0 yy yy ∂y2 y y +Nˆe´u z = f(x, y) c´oc´acd¯a.o h`amriˆengcˆa´p 2 liˆentu. c trong miˆe` n D th`ıtrong miˆe` n 00 00 d¯´o: zxy = zyx. *Nˆe´u z = f(u, v) l`ah`amkha’ vi v`a u = u(x, y),v = v(x, y) c´oc´acd¯a.o h`amriˆeng 0 0 0 0 ux,uy,vx,vy trong miˆe` n D th`ıtrong miˆ`end¯´otˆo`nta.i c´acd¯a.o h`amriˆeng 0 0 0 0 0 zx = zu · ux + zv · vx; 0 0 0 0 0 zy = zu · uy + zv · vy
- 27 u . 2 2 0 0 V´ıdu Cho z = e sin v v´oi u = xy, v = x + y .T´ınh zx,zy. 0 u 0 u 0 0 0 0 V`ı: zu = e sin v; zv = e cos v; ux = y; uy = x; vx =2x; vy =2y, nˆen: 0 0 0 0 0 u u xy 2 2 xy 2 2 + zx = zu ·ux +zv ·vx = e sin v·y+e cos v·2x = ye sin(x +y )+2xe cos(x +y ) 0 0 0 0 0 u u xy 2 2 xy 2 2 + zy = zu ·uy +zv ·vy = e sin v·x+e cos v·2y = xe sin(x +y )+2ye cos(x +y ) . 1.3. Cu. c tri. cu’a h`amhai biˆe´n . * Cho h`am z = f(x, y) x´acd¯i.nh, liˆentu. c trong miˆ`en D. Ta n´oi z d¯ a.t cu. cd¯a.i . . . . . . . (tuo ng tu. ,cu. ctiˆe’u) d¯i.a phuo ng ta.i Mo(xo,yo) ∈ D nˆe´utˆo`nta.i khoa’ng ho’ U . . . cu’a Mo(xo,yo) trong D sao cho f(xo,yo) ≥ f(x, y) (tuong tu. , f(xo,yo) ≤ f(x, y)) . v´oimo.i(x, y) ∈ D. . . + Quy t˘a´ct`ım cu. c tri.: Gia’ su’ z = f(x, y) c´od¯a.o h`amriˆeng liˆentu. cd¯ˆe´ncˆa´p . . 0 0 - 2 trong khoa’ng ho’ ch´ua Mo(xo,yo) v`ac´o fx(xo,yo)=fy(xo,yo)=0.D˘a.t A = 00 00 00 fxx(xo,yo),B = fxy(xo,yo),C = fyy(xo,yo), th`ı: 2 . +Nˆe´u B − AC 0th`ız = f(x, y)d¯a.tcu. ctiˆe’uta.i(xo,yo); 2 . +Nˆe´u B − AC > 0th`ı(xo,yo) khˆongpha’i l`ad¯iˆe’mcu. c tri.; 2 . . +Nˆe´u B − AC = 0 th`ıkhˆongkˆe´t luˆa.nd¯uo. c. . V´ıdu T`ım cu. c tri. cu’a h`amsˆo´: a. z = f(x, y)=x2 − xy + y2 +3x − 2y +1 b. z = x3 + y3 − 3xy 0 0 00 00 00 a. Ta c´o: zx =2x − y +3; zy = −x +2y +2; zxx =2; zxy = −1; zyy =2. 4 0 x = − zx =0 2x − y +3 =0 Gia’i ⇔ ⇔ 3 z0 =0 −x +2y − 2=0 1 y y = 3 4 1 ’ − 00 00 − 00 Ta.id¯iˆem Mo , , ta c´o: A = zxx =2,B = zxy = 1,C = zyy =2, 3 3 Mo Mo Mo nˆen: B2 − AC =(−1)2 − 2 · 2=−3 0. Vˆa.y Mo(0, 0) khˆongpha’i l`acu. c tri ’ 00 00 − 00 Ta.id¯iˆem M1(1, 1), ta c´o: A = zxx =6,B = zxy = 3,C = zyy =6, Mo Mo Mo 2 . . nˆen: B − AC =9− 36 = −27 < 0. Suy ra h`am2 biˆe´nd¯a.tcu. ctiˆe’uta.i M1(1, 1) v´oi zmin = −1. BAI` TAˆ. P x − y 3.1.1. Cho h`am f(x, y)= .Ch´u.ng minh: x + y lim limf(x, y) = 1; lim limf(x, y) = −1 x→0 y→0 y→0 x→0
- 28 trong khi d¯´o limf(x, y) khˆongtˆo`nta.i. x→0 y→0 x2y2 3.1.2. Cho h`am f(x, y)= .Ch´u.ng minh: x2y2 +(x − y)2 lim limf(x, y) = lim limf(x, y) =0 x→0 y→0 y→0 x→0 , trong khi d¯´o limf(x, y) khˆongtˆo`nta.i. x→0 y→0 1 1 3.1.3. Cho h`am f(x, y)=(x + y) sin sin .Ch´u.ng minh lim limf(x, y) v`a x y x→0 y→0 lim limf(x, y) khˆongtˆo`ntai, nhu.ng limf(x, y)=0 y→0 x→0 . x→0 y→0 . 3.1.4. T´ınh c´acgi´oiha.n sau: 2 2 2 x x + y x + y xy 2 2 lim ; lim ; lim ; lim(x2 + y2)x y x→0 x2 − xy + y2 x→0 x4 + y4 x→0 x2 + y2 x→0 y→0 y→0 y→0 y→0 3.1.5. Cho h`am x2 − y2 xy nˆe´u x2 + y2 =06 f(x, y)= x2 + y2 0 nˆe´u x2 + y2 =0. . Ch´ung minh f”yx(0, 0) =6 f”xy(0, 0). . . . . 3.1.6. Nghiˆen c´uucu. c tri. d¯ i .a phuong cu’a c´ach`amsau: a. z = x4 + y4 − x2 − 2xy − y2 +1 b. x =2x4 + y4 − x2 − 2y2 − 1 II. T´ıch phˆanhai l´o.p 1. D- `ınhngh˜ıa, t´ınh chˆa´t . . . . . Xuˆa´t ph´att`u c´acb`aito´anthu. ctˆe´ (nhu t´ınhthˆe’ t´ıch vˆa.tthˆe’ h`ınhtru. ,d¯u`ong k´ınh mˆo.tmiˆ`en), ta c´od¯i.nh ngh˜ıasau: . * Cho h`am z = f(x, y) x´acd¯i.nh trong miˆe` nh˜uuha.n D trong xOy. Phˆanhoa.ch D th`anh n miˆe` n nho’ tu`y´yc´otˆenv`adiˆe.nt´ıch∆s1, ∆s2, ,∆sn.Trˆen mˆo˜i∆Si (i =1, ,n), lˆa´y Mi(xi,yi) tu`y´yv`ago.itˆo’ng n In = Xf(xi,yi)∆si i=1 l`a tˆo’ng t´ıch phˆancu’a f(x, y) trong D. . . . Nˆe´u khi d¯u`ong k´ınh l´on nhˆa´tcu’a c´acmiˆe` n∆si dˆa`nd¯ˆe´n 0 (max di → 0) m`a In dˆa`n . d¯ ˆe´nmˆo.t gi´oiha.n x´acd¯i.nh I, khˆongphu. thuˆo.c c´ach chia miˆe` n D (phˆanhoa.ch) v`a
- 29 . . . c´ach cho.n Mi(xi,yi) trong mˆo˜imiˆe` n∆si th`ıgi´oiha.n d¯ ´o d¯ u o. cgo.il`at´ıch phˆan . hai l´o pcu’a f(x, y) trong miˆe` n D v`ak´yhiˆe.u l`a: n ZZ f(x, y)ds = lim Xf(xi,yi)∆si max di→0 D i=1 trong d¯´o f(x, y)l`ah`amdu.´o .idˆa´ut´ıch phˆan, D l`a miˆ`enlˆa´yt´ıch phˆan, ds l`a yˆe´utˆo´ diˆe.nt´ıch, x, y l`a biˆe´nt´ıch phˆan. . . . . . . . + Khi t´ıch phˆanhai l´optˆo`nta.i, ta c´othˆe’ chia D bo’ ilu´oi c´acd¯u`o ng song song v´oi . Ox, Oy, khi d¯´o∆si l`ah`ınhch˜u nhˆa.t, yˆe´utˆo´ diˆe.nt´ıchds b˘a`ng dx, dy: n ZZ f(x, y)dxdy = lim Xf(xi ,yi)∆si max di→0 D i=1 . . +Diˆe.n t´ıch miˆ`en D d¯ u o. c t´ınhb˘a`ng: S(D)=ZZ dxdy D . . +Tˆo’ ho. p tuyˆe´n t´ınhnh˜ung h`amkha’ t´ıch trˆen D c˜ung kha’ t´ıch trˆen D v`a: ZZ ZZ ZZ [αf1(x, y) ± βf2(x, y)]dxdy = α f1(x, y)dxdy ± β f2(x, y)dxdy D D D +Nˆe´u f(x, y) kha’ t´ıch trˆen D th`ı |f(x, y)| c˜ung kha’ t´ıch trˆen D v`a: ZZ ZZ f(x, y)dxdy ≤ |f(x, y)|dxdy D D . . . . + Chia D th`anh2 miˆ`en D1,D2 r`oi nhau bo’ imˆo.td¯u`ong L.Nˆe´u f(x, y) kha’ t´ıch trˆen ca’ D1,D2 (kˆe’ ca’ biˆen L) th`ın´okha’ t´ınhtrˆen D v`a: ZZ f(x, y)dxdy = ZZ f(x, y)dxdy + ZZ f(x, y)dxdy D D1 D +Nˆe´u f(x, y) kha’ t´ıch trˆen D v`a m ≤ f(x, y) ≤ M,∀(x, y) ∈ D, th`ı: ZZ f(x, y)dxdy ∃µ ∈ [m, M]: µ = D S(D)
- 30 +Nˆe´u f(x, y),g(x, y) kha’ t´ıch trˆen D v`athoa’ m˜an f(x) ≤ g(x) th`ı: ZZ f(x, y)dxdy ≤ ZZ g(x, y)dxdy D D 2. C´acht´ınh t´ıch phˆanhai l´o.p +Nˆe´u h`amsˆo´ f(x, y)liˆen tu. ctrˆen miˆe` n D = {a ≤ x ≤ b; ϕ(x) ≤ y ≤ ψ(x)} trong d¯´o ϕ(x)v`aψ(x)liˆen tu. ctrˆen [a, b] th`ı: b " ϕ(x) # ZZ f(x, y)dxdy = Z Z f(x, y)dy dx a ψ(x) D . . V´ıdu. 1. T´ınh thˆe’ t´ıch h`ınhtru. gi´oiha.nbo’ i c´acm˘a.t: x =0,x=1,y= −1,y=1,z=0,z= x2 + y2. 1 1 1 2 4 Ta c´o: V = ZZ f(x, y)dxdy = Z Z (x2 + y2)dy dx = Z 2x2 + dx = 0 −1 0 3 3 D . . V´ıdu. 2. T´ınh thˆe’ t´ıch h`ınhtru. gi´oiha.nbo’ im˘a.t 2 z = f(x, y)=xy , m˘a.t z =0,x=0,x=1,y= −2,y=3. 1 3 2 1 3 3 Z Z 2 x y 35 Ta c´o: V = xdx · y dy = = 0 −2 2 0 3 −2 6 . . V´ıdu. 3. T´ınh thˆe’ t´ıch h`ınhtru. gi´oiha.nbo’ i c´acm˘a.t: x2 x =1,x=2,y= ,y= x2,z=0,z= xy. 2 2 2 " x # 2 3x5 63 Ta c´o: V = ZZ xydxdy = Z Z xydy dx = Z dx = . 2 1 x 1 8 16 D 2 ZZ . . . . . 2 V´ıdu. 4. T´ınh xdxdy,v´oi D l`amiˆe` n gi´oiha.nbo’ i c´acd¯u`ong y = x v`a y = x . D . . 2 Miˆe` n D d¯ u o. c x´acd¯i.nh: D = {0 ≤ x ≤ 1; x ≤ y ≤ x}, nˆen: 1 x 1 ZZ xdxdy = Z Z xdy dx = . 0 x2 12 D ZZ . . . . . . V´ıdu. 5. T´ınh I = (x − y)dxdy, trong d¯´o D d¯ u o. c gi´oiha.nbo’ i c´acd¯u`o ng D y = ±1,x= y2,y= x +1.
- 31 . . 2 Miˆe` n D d¯ u o. c x´acd¯i.nh: D = {−1 ≤ y ≤ 1; y − 1 ≤ x ≤ y }, suy ra: 2 1 " y # 1 y4 y2 1 −7 I = Z Z (x − y)dx dy = Z − y3 + − dy = . −1 y−1 −1 2 2 2 15 0 x = x(u, v) + Cho f(x, y) liˆentu. c trˆen D d¯´ongv`abi. ch˘a.n, l`aa’nh cu’a D qua ´anhxa. . y = y(u, v) Nˆe´u x(u, v),y(u, v)liˆen tu. c v`ac´oc´acd¯a.o h`amriˆeng liˆen tu. c ∂x ∂x J(u, v)= ∂u ∂v =06 , ∀(u, v) ∈ D0 ∂y ∂y ∂u ∂v th`ı: ZZ f(x, y)dxdy = ZZ f[x(u, v),y(u, v)]|J(u, v)|dudv D D0 x = r cos ϕ Ch˘a’ ng ha.n khi d¯˘a.t th`ı: y = r sin ϕ cos ϕ −r sin ϕ J(u, v)= = r sin ϕrcos ϕ khi d¯´o: ZZ f(x, y)dxdy = ZZ f(r cos ϕ, r sin ϕ)rdrdϕ D D0 ZZ −x2−y2 . . . V´ıdu. 6. T´ınh t´ıch phˆan I = e dxdy trong d¯´o D l`ad¯u`o ng tr`ond¯onvi D 2π 1 2π 2 1 1 1 Ta c´o: I = Z dϕ Z e−r rdr = Z 1 − dϕ = π 1 − . 0 0 0 2 e e ZZ . V´ıdu. 7. T´ınh t´ıch phˆan I = (x +2y)dxdy, trong d¯´o D l`ah`ınhb`ınh h`anhgi´oiha.n D bo’.i c´acd¯u.`o.ng x + y =1,x+ y =2, 2x − y =1, 2x − y =3. 1 1 1 Ta c´o: D0 = {(u, v): 1≤ u ≤ 2, 1 ≤ v ≤ 3} v`a J = 3 3 = − =06 , nˆen: 2 1 3 − 3 3 1 u + v 4u − 3v I = ZZ − + dudv 3 3 3 D0 1 2 3 1 2 11 = − Z Z (5u − v)dv du = − Z (10u − 4)du = − 9 1 1 9 1 9
- 32 ZZ . V´ıdu. 8. T´ınh I = ydxdy v´oi D l`amiˆe` n: D . . a. H`ınh qua.t tr`ontˆam O, b´ank´ınh a n˘a`m trong g´ocphˆa`ntuth´u 2. . . . . . . . b. Miˆe` n gi´oiha.nbo’ i c´acd¯u`ong cong c´ophuong tr`ınhtrong hˆe. toa. d¯ ˆo. cu. c l`a: r = 2 + cos ϕ, r =1. π a a3 π a3 a. I = Z sin ϕdϕ Z r2dr = Z sin ϕdϕ = . π 3 π 3 2 0 2 2π 2+cos ϕ 1 2π b. I = Z Z rdr sin ϕdϕ = Z (3 + 4 cos ϕ + cos2 ϕ) sin ϕdϕ =0. 0 1 2 0 ZZ V´ıdu. 9. T´ınh I = p4 − x2 − y2dxdy trong d¯´o: D D l`anu’.atrˆen cu’a h`ınhtr`on(x − 1)2 + y2 ≤ 1. x = r cos ϕ 0 π D- ˘a.t th`ı D = {(r, ϕ): 0≤ r ≤ 2 cos ϕ, 0 ≤ ϕ ≤ }, khi d¯´o: y = r sin ϕ 2 π 2 cos ϕ π 2 8 2 8 π 2 I = Z Z p4 − r2rdr dϕ = Z (1 − sin3 ϕ)dϕ = − . 0 0 3 0 3 2 3 III. T´ıch phˆan3 l´o.p 1. D- .inh ngh˜ıa, t´ınh chˆa´t . . 3 Cho f l`amˆo.t h`ambi. ch˘a.n, x´acd¯i.nh trˆen mˆo.ttˆa.p V d¯od¯uo. c trong R . Chia V . . . . th`anhh˜uuha.nnh˜ung tˆa.p Vi d¯od¯uo. c, khˆongc´od¯iˆe’m trong chung. Lˆa.ptˆo’ng t´ıch phˆan n X f(ξ,η,τ)∆Vi (1) i=1 . o’ d¯ˆay∆Vi l`athˆe’ t´ıch tˆa.p Vi,v`a(ξ,η,τ) l`amˆo.td¯iˆe’m tu`y´ythuˆo.c Vi. . . . *Go.i D l`asˆo´ l´on nhˆa´t trong c´acd¯u`ong k´ınh d(Vi)cu’a ph´epphˆanhoa.ch {Vi}1≤i≤n. Nˆe´u n lim X f(ξ,η,τ)∆Vi D→0 i=1 . . . tˆo`nta.i th`ıgi´atri. n`ayd¯uo. cgo.il`at´ıch phˆanba l´o pcu’a h`am f trˆentˆa.p V v`a . . d¯ u o. ck´yhiˆe.ul`a ZZZ f(x, y, z)dxdydz, V . . h`am f d¯ u o. cgo.il`akha’ t´ınh trˆen V . . . . . . . +T´ıch phˆanba l´op c´oc´act´ınh chˆa´t ho`anto`antuong tu. nhu t´ıch phˆanhai l´op. 2. C´acht´ınh t´ıch phˆanba l´o.p +Nˆe´umiˆe` nlˆa´y t´ıch phˆanl`amˆo.th`ınh hˆo.p V =[a1,b1] × [a2,b2] × [a3,b3],
- 33 th`ı: b1 " b2 b3 ! # ZZZ f(x, y, z)dxdydz = Z Z Z f(x, y, z)dz dy dx a1 a2 a3 V ZZZ . V´ıdu T´ınh I = xyzdxdydz v´oi V =[0, 1] × [2, 4] × [5, 8]. V 1 4 8 2 1 2 4 2 8 Z Z Z x y z 1 39 117 I = xdx · ydy · zdz = · · = · 6 · = 0 2 5 2 0 2 2 2 5 2 2 2 . . . +Nˆe´umiˆe` n l`amˆo.tthˆe’ tru. mo’ rˆo.ng (gi´oiha.nbo’ i2m˘a.t ψ1(x, y),ψ2 (x, y), m˘a.t tru. c´od¯u.`o.ng sinh song song Oz,d¯u.`o .ng chuˆa’n l`abiˆen Dxy = {(x, y): a ≤ x ≤ y,ϕ1(x) ≤ y ≤ ϕ2(x)} . v´oi ϕ1,ϕ2 liˆentu. ctrˆen [a, b] th`ı: b " ϕ2(x) ψ1(x,y) ! # ZZZ f(x, y, z)dxdydz = Z Z Z d(x, y, z)dz dy dx a ϕ1(x) ψ2(x,y) V ZZZ . . . V´ıdu. . T´ınh (1 − x − y)dxdydz v´oimiˆe` n V gi´oiha.nbo’ i c´acm˘a.t ph˘a’ ng toa. d¯ ˆo. v`a V m˘a.t ph˘a’ ng x + y + z =1. Ta c´o: V = {(x, y, z): 0≤ x ≤ 1, 0 ≤ y ≤ 1 − x, 0 ≤ z ≤ 1 − x − y}, nˆen: 1 1−x 1−x−y 1 1−x I = Z Z Z (1 − x − y)dz dy dx = Z Z (1 − x − y)2dy dx 0 0 0 0 0 1 1 1 = Z (1 − x)3(1 − x3)dx = 0 3 12 . . . * D- ˆo’ibiˆe´n trong t´ıch phˆanba l´o p: Cho f liˆentu.ctrˆen miˆe` n d¯´ong,d¯od¯uo. cv`a x = x(u, v, w) 3 . 0 . bi. chˆa.n V ⊂ R ,v´oi V l`aa’nh cu’a V qua d¯on ´anh y = y(u, v, w) Nˆe´u c´ach`am z = z(u, v, w). sˆo´ x = x(u, v, w),y = y(u, v, w),z = z(u, v, w) liˆentu.c, c´oc´acd¯a.o h`amriˆeng liˆen 0 tu.ctrˆen V v`anˆe´u x0 x0 x0 D(x, y, z) u v w J(u, v, w)= = y0 y0 y0 =06 D(u, v, w) u v w z0 z0 z0 u v w
- 34 0 th`ıta.imo.id¯iˆe’m (u, v, w) ∈ V , ta c´o: ZZZ f(x, y, z)dxdydz = ZZZ f[x(u, v, w),y(u, v, w),z(u, v, w)]|J(u, v, w)|dudvdw V V 0 x = r cos ϕ +Nˆe´u theo toa. d¯ ˆo. tru.: y = r sin ϕ th`ı: z = r, cos ϕ −r sin ϕ 0 J(r, ϕ, z)= sin ϕrcos ϕ 0 001 nˆen ZZZ f(x, y, z)dxdydz = ZZZ f(r cos ϕ, r sin ϕ, z)rdrdϕdz V V 0 ZZZ 2 2 . . V´ıdu T´ınh I = (x + y )zdxdydz trong d¯´o V l`amiˆ`en gi´oiha.nbo’ i c´acm˘a.t V x2 + y2 =1v`az =2. . . . H`ınh tru. tr`onxoay V d¯ u o. c x´acd¯i.nh bo’ i: V = {(r, ϕ, z): 0≤ r ≤ 1, 0 ≤ ϕ ≤ 2π,0 ≤ z ≤ 2}, suy ra: 2 2π 1 I = Z zdz · Z dϕ · Z dr = π. 0 0 0 x = r cos ϕ sin θ +Nˆe´u theo toa. d¯ ˆo. cˆa`u: y = r sin ϕ sin θ th`ı: z = r cos θ, cos ϕ sin θrcos ϕ cos θ −r sin ϕ sin θ J(r, ϕ, θ)== sin ϕ sin θrcos θ sin ϕrcos ϕ sin θ cos θ −r sin θ 0 nˆen ZZZ f(x, y, z)dxdydz == ZZZ f(r cos ϕ sin θ, r sin ϕ sin θ, r cos θ)r2 sin θdrdϕdθ V V 0 ZZZ 2 2 . . V´ıdu T´ınh I = (x + y )dxdydz trong d¯ˆo´ V l`amiˆ`en gi´oiha.nbo’ im˘a.tcˆa`u V x2 + y2 + z2 =1v`am˘a.t n´on x2 = y2 − z2 =0(z>0).
- 35 √ 2 2 2 2! 2 2 2 x + y = x + y + z =1 . . 2 Gia’ihˆe. giao tuyˆe´n l`ad¯u`ong tr`on x2 + y2 − z2 =0, √ 2 z = suy ra: 2 π V = {(r, ϕ, theta): 0≤ r ≤ 1, 0 ≤ ϕ ≤ 2π,0 ≤ θ ≤ } 4 v`a f(x, y, z)=x2 + y2 = r2 sin θ,nˆen 1 2π π π 4 1 4 I = Z r4dr · Z dϕ · Z dθ = 2π Z (1 − cos2 θ)d(− cos θ) 5 0 0 0 √ 0 3 π 4 − 2π cos θ 8 5 2 = − cos θ = π 5 3 0 30 ´. * U ng du.ng cu’at´ıch phˆank´ep . . 2 +Diˆe.n t´ıch cu’amˆo.th`ınh ph˘a’ ng D d¯´ong,d¯od¯uo. c, bi. chˆa.n trong R S(D)=ZZ dxdy D . . +Thˆe’ t´ıch miˆe` n V d¯od¯uo. c, d¯´ong,bi. chˆa.n V = ZZZ dxdydz V Nˆe´u V l`ah`ınh tru. cong, x´et D l`ah`ınh chiˆe´ucu’a V xuˆo´ng m˘a.t ph˘a’ ng, z = f(x, y) l`am˘a.ttrˆen h`ınhtru. cong: V = ZZ f(x, y)dxdy D . Nˆe´u V l`athˆe’ tru. mo’ rˆo.ng, x´et D l`ah`ınh chiˆe´ucu’a D lˆen xOy, z = . . ψ1(x, y),z = ψ2(x, y) l`am˘a.tdu´oi, m˘a.ttrˆen cu’a V : ZZZ V = f(ψ1(x, y),ψ2 (x, y))dxdy D . . . . . . V´ıdu. 1. T´ınh diˆe.nt´ıchh`ınh gi´oiha.nbo’ i c´acd¯u`ong th˘a’ ng x =1,x= 2 v`ac´acd¯u`ong a2 2a2 y = ,y = (x>0) x x
- 36 a2 2a2 D = {(x, y): 1≤ x ≤ 2, ≤ y ≤ }, suy ra: x x 2 2 2a 2 " x # 2a2 a2 S(D)=ZZ dxdy = Z Z dx dx = Z − dx = a2 ln 2 2 1 a 1 x x D x . . V´ıdu. 2. T´ınh thˆe’ t´ıch vˆa.tthˆe’ V gi´oiha.nbo’ i c´acm˘a.t x2 + y2 =2,z=4− x2 − y2,z=0. . . 2 2 V l`ah`ınhtru. cong, m˘a.ttrˆen c´ophuong tr`ınh z =4− x − y , h`ınhchiˆe´u D cu’a V lˆenm˘a.t ph˘a’ ng xOy l`ah`ınhtr`on x2 + y2 ≤ 2. Vˆa.y √ 2π 2 V = ZZ (4 − x2 − y2)dxdy = Z · Z (4 − r2)rdr =6π 0 0 D . . + Cho S l`am˘a.t cong c´ophuong tr`ınh z = f(x, y), trong d¯´o f liˆentu. c, c´od¯a.o h`am . . riˆeng liˆen tu. ctrˆen miˆe` n d¯´ong,bi. chˆa.n, d¯od¯uo. c, th`ıdiˆe.n t´ıch m˘a.t cong S l`a: ZZ q 0 2 0 2 S = 1+fx + fy dxdy D V´ıdu T´ınh diˆe.n t´ıch phˆa`nm˘a.tcˆa`u x2 + y2 + z2 = a2 n˘a`m trong m˘a.t tru. x2 + y2 = a2. . M˘a.t tru. c˘a´tm˘a.tcˆa`u th`anhhai ma’nh d¯ˆo´ix´ung nhau qua m˘a.t ph˘a’ ng xOy,mˆo˜i . . . ma’nh n`ayla.id¯uo. c c´acm˘a.t ph˘a’ ng toa. d¯ ˆo. chia th`anh4 ma’nh b˘a`ng nhau. V´oi z ≥ 0, ta a2 c´o: z = pa2 − x2 − y2, suy ra 1 + z0 2 + z0 2 = ,nˆen x y a2 − x2 = y2 2π a ZZ a Z Z rdr 2 S =8 dxdy =8x dϕ0 · √ =4πa 2 2 pa2 − x2 − y2 0 a − r x2+y2≤a2,x≥0,y≥0 BAI` TAˆ. P 3.2.1. T´ınh ZZ x ln ydxdy D . . v´oi D l`ah`ınhch˜u nhˆa.t: 0 ≤ x ≤ 4, 1 ≤ y ≤ e. 3.2.2. T´ınh ZZ (cos2 x + sin2 y)dxdy D
- 37 v´o.i D l`ah`ınhvuˆong: π π 0 ≤ x ≤ , 0 ≤ y ≤ . 4 4 3.2.3. T´ınh 2 " x2 # I = Z Z (2x − y)dy dx. 1 x 3.2.4. T´ınh ZZ (x − y)dxdy D . . . v´oi D l`ah`ınhgi´oiha.nbo’ i: y =2− x2,y=2x − 1. 3.2.5. T´ınh ZZ (x +2y)dxdy D . . . . . v´oi D l`ah`ınhgi´oiha.nbo’ i c´acd¯u`o ng th˘a’ ng: y = x, y =2x, x =2,x=3. 3.2.6. T´ınh ZZ ex+sin y cos ydxdy D . . v´oi D l`ah`ınhch˜u nhˆa.t: π 0 ≤ x ≤ π, 1 ≤ y ≤ . 2 3.2.7. T´ınh ZZ (x2 + y2)dxdy D . . . . . v´oi D l`amiˆe` n gi´oiha.nbo’ i c´acd¯u`ong y = x, x =0,y=1,y=2. 3.2.8. T´ınh ZZ ln(x2 + y2)dxdy D v´o.i D l`amiˆe` n h`ınhv`anhkh˘angi´u.x hai d¯u.`o .ng tr`on x2 + y2 = e2 v`a x2 + y2 = e4.
- 38 3.2.9. T´ınh ZZ (x2 + y2)dxdy D v´o.imiˆ`en D gi´o.i h`anbo’.id¯u.`o.ng tr`on x2 + y2 =2ax. 3.2.10. T´ınh ZZ x3ydxdy D . . . . . v´oi D l`amiˆe` n gi´oiha.nbo’ i c´acd¯u`ong y =0v`a y = p2ax − x2. 3.2.11. T´ınh ZZ sin(x + y)dxdy D . . . . . v´oi D l`amiˆe` n gi´oiha.nbo’ i c´acd¯u`ong π y =0,y= x, x + y = . 2 3.2.12. T´ınh ZZ x2(y − x)dxdy D . . . . . v´oi D l`amiˆe` n gi´oiha.nbo’ i c´acd¯u`ong x = y2 v`a y = x2. 3.2.13. T´ınh ZZ f(x, y)dxdy D . . . . . v´oi D l`amiˆe` n gi´oiha.nbo’ id¯u`o ng x2 y2 + =1, a2 b2 c`onh`amdu.´o.idˆa´u t´ıchphˆan 2 2 cq1− x − y a2 b2 f(x, y)=Z tdt. 0 3.2.14. T´ınh ZZ r2drdϕ D
- 39 v´o.i D l`amiˆe` n: a. C´acd¯u.`o.ng tr`on r = a v`a r =2a. b. D- u.`o .ng r = a sin 2ϕ. 3.2.15. T´ınh ZZ r sin ϕdrdϕ D v´o.i D l`amiˆe` n: π a. Quat tr`ongi´o.ihanbo’.i c´acd¯u.`o.ng r = a, ϕ = ,ϕ= π. . . 2 π b. Nu’.ad¯u.`o.ng tr`on r ≤ 2a cos ϕ, 0 ≤ ϕ ≤ . 2 c. Nu’.ad¯u.`o.ng tr`on r = 2 + cos ϕ v`a r =1. . . ’ . 3.2.16. Su’ du√.ng cˆongth´ucd¯ˆoibiˆe´n trong toa. d¯ ˆo. cu. c, t´ınhc´act´ıch phˆan: R " R2−x2 # a. Z Z ln(1 + x2 + y2)dy dx 0 0 √ R " Rx−x2 # Z Z 2 2 2 b. √ pR − x − y dy dx 0 − Rx−x2 3.2.17. a. T´ınh 1 2x Z Z dy dx 0 x x = u(1 − v) b˘a`ng c´achd`ung c´acbiˆe´nm´o.i y = uv b. T´ınh ZZ dxdy D . . . . nˆe´u D gi´oiha.nbo’ i c´acd¯u`ong xy =1,xy=2,y= x, y =3x. 3.2.18. T´ınh c´act´ınhphˆanba l´o.p sau: x2 y2 z2 x2 y2 z2 a. I = ZZZ + + dxdydz v´o.i V gi´o.ihanbo’.im˘at + + =1. a2 b2 c2 . . a2 b2 c2 V ZZZ 2 2 . . . . . 2 2 2 b. I = (x + y )dxdydz,v´oi V d¯ u o. c gi´oiha.nbo’ i c´acm˘a.t x + y = z ,z=2. V ZZZ 2 2 . . . . . 2 2 c. I = (x + y )dxdydz,v´oi V d¯ u o. c gi´oiha.nbo’ i c´acm˘a.t x + y =2z, z =2. V
- 40 ZZZ . . . 3.2.19. T´ınh I = xyzdxdydz,v´oi V n˘a`m trong g´ocphˆa` n t´amth´u nhˆa´t, gi´oiha.n V . . bo’ i c´acm˘a.t sau, v´oi 0 <a<b,0 <α<β,0 <m<n: x2 + y2 x2 + y2 z = ,z = ,xy = a2,xy = b2,y = αx, y = βx m n -ooOoo-
- 41 Chu.o.ng 4 . . PHUONG TR`INH VI PHANˆ I. Phu.o.ng tr`ınhvi phˆancˆa´p1 1. Kh´ainiˆe.m chung . . . . *Tago.i phuo ng tr`ınh vi phˆancˆa´p1l`aphuong tr`ınhc´oda.ng F (x, y, y0)=0 (I) ho˘a.c 0 y = f(x, y)(Io) 0 trong d¯´o x l`abiˆe´nsˆo´, y l`ah`amcu’a x,v`ay l`ad¯a.o h`amcu’a y. . . . . *Nˆe´u c´oh`am y = ψ(x) tho’a m˜anphuong tr`ınh(I)hay(Io)th`ıy = ψ(x)d¯uo. cgo.i . . l`a nghiˆe.mcu’a phuo ng tr`ınh (I) hay (Io). . . *Nˆe´u c´oh`am y = ψ(x, C) ho˘a.chˆe. th´ucΦ(x, y, C)=0tho’a m˜an(I)hay(Io)v´oi C . . t`uy´ytrong miˆe` n n`aod¯´ocu’a R, v`av´oimˆo˜id¯iˆ`eukiˆe.nd¯ˆa`u y(xo)=yo v´oi(xo,yo) . . thuˆo.cmiˆe` n x´acd¯i.nh cu’aphuong tr`ınh,chı’ c´oduy nhˆa´t gi´atri. C = Co l`amcho y = ψ(x, Co) hay Φ(x, y, Co)=0tho’a m˜and¯iˆe` ukiˆe.nd¯ˆa`u, th`ı y = ψ(x, C) ho˘a.c . . . . Φ(x, y, C)=0d¯uo. cgo.il`anghiˆe.mtˆo’ng qu´atcu’a phuo ng tr`ınh (I) hay (Io). *Nˆe´u y = ψ(x, C) hay Φ(x, y, C) = 0 l`anghiˆe.mtˆo’ng qu´atcu’a(I)hay(Io), cho . . C = Co (gi´atri. cu. thˆe’ x´acd¯i.nh) th`ı y = ψ(x, Co) hay Φ(x, y, Co)=0d¯uo. cgo.i l`a nghiˆe.m riˆengcu’a (I) hay (Io). Nˆe´u nghiˆe.m y = ψ(x) khˆongpha’i l`anghiˆe.m . . riˆeng nhˆa.nt`unghiˆe.mtˆo’ng qu´atv´oibˆa´tk`y gi´atri. C n`ao(kˆe’ ca’ C = ±∞) th`ıta go.i n´ol`a nghiˆe.mk`ydi. cu’a (I) hay (Io). . . + (D- .inh l´ytˆo`nta.i v`aduy nhˆa´t nghiˆe.m): Cho phuong tr`ınh (Io).Nˆe´u f(x, y) . liˆentu.c trong miˆ`en n`aod¯´och´uad¯iˆe’m (xo,yo) th`ı tˆo`nta.i ´ıt nhˆa´tmˆo.t nghiˆe.m 0 y = ψ(x) sao cho yo = ψ(xo) v`anˆe´u fy (x, y) liˆentu.cta.i (xo,yo) th`ı y = ψ(x) tˆo`n ta.i duy nhˆa´t. . . 2. C´acloa.i phuo ng tr`ınh vi phˆancˆa´p1 2.1. Phu.o.ng tr`ınhbiˆe´nsˆo´ phˆanly dy L`aphu.o.ng tr`ınhm`anˆe´u thay y0 = th`ıc´othˆe’ biˆe´nd¯ˆo’ivˆ`e dang f (y)dy = dx . 1 . . . . f2(x)dx.Lˆa´y t´ıch phˆanbˆa´td¯i.nh2vˆe´ th`ıgia’id¯uo. cphuong tr`ınh. . . V´ıdu. 1. Gia’iphuong tr`ınh: ydy =(x2 +1)dx. Lˆa´y t´ıch phˆanhai vˆe´ cu’aphu.o.ng tr`ınh d¯˜acho: y2 x3 C 2 Z ydy = Z (x2 +1)dx ⇔ = + x + ⇔ y2 = x3 +2x + C 2 3 2 3 . . V´ıdu. 2. Gia’iphuong tr`ınh: (y − x2y)dy +(xy2 + x)dx =0.tag1 Ta c´o: (1)⇔ y(x2 − 1)dy = x(y2 +1)dx (2)
- 42 +Nˆe´u x2 − 1 ≡ 0 ⇔ x ≡±1th`ıdx = 0, nˆen(2) tho’a m˜an.Vˆa.y x = ±1 l`anghiˆe.m cu’a (1). y x +Nˆe´u x2 − 1 6≡ 0 ⇔ x 6≡ ±1: (2)⇔ dy == dx.Lˆa´y t´ıch phˆan2 vˆe´: y2 +1 x2 − 1 ydy xdx 1 1 1 Z = Z ⇔ ln |y2 +1| = ln |x2 − 1| + ln |C| y2 +1 x2 − 1 2 2 2 " y2 +1=C(x2 − 1), ∀C =06 ⇔ y2 +1=C(x2 − 1) (∀C =6 0). Vˆa.y (1) c´onghiˆe.m: x = ±1 . . V´ıdu. 3. Gia’iphuong tr`ınh: y0 =3x2y (1) dy Ta c´o: (1)⇔ =3x2y ⇔ dy =3x2ydx (2) dx 0 +Nˆe´u y ≡ 0th`ıy = 0, nˆen(2) tho’a m˜an.Vˆa.y y = 0 l`anghiˆe.mcu’a (1). dy +Nˆe´u y 6≡ 0: (2)⇔ =3x2dx.Lˆa´y t´ıch phˆan2 vˆe´: y 3 3 ln |y| = x3 +ln|C|⇔ln |y| =ln|Cex |⇔y = Cex , ∀C =06 x3 . Vˆa.y (1) c´onghiˆe.m: y = Ce (v´oi C t`uy´y). 2.2. Phu.o.ng tr`ınhvi phˆand¯˘a’ ng cˆa´pcˆa´p1 . . 0 . L`aphuong tr`ınhc´oda.ng y = f(x, y)v´oi f(λx, λy)=f(x, y), ∀λ =0.6 0 0 . . . D- ˘a.t y = ux, ta c´o: u x + u = y = f(x, y)=g(u), ta d¯uavˆe` phuong tr`ınhc´obiˆe´n sˆo´ phˆanly u0x = g(u) − u. . . V´ıdu. 6. Gia’iphuong tr`ınh: x + y y0 = (1) x − y 0 0 D- ˘a.t y = ux ⇒ y = u x + u, ta c´o: x + ux 1+u 1 − u dx (1)⇔ u0x + u = ⇔ u0x = − u ⇔ du = . x − ux 1 − u 1+u2 x Lˆa´y t´ıch phˆan2 vˆe´: du 1 2udu 1 ln |1+u2| ln |Cx2| Z − Z =ln|x| + ln |C|⇔arctg u − = 1+u2 2 1+u2 2 2 2 Vˆa.y (1) c´onghiˆe.m: y 2Arctg =ln|C(x2 + y2)|, ∀C =06 x . . V´ıdu. 5. Gia’iphuong tr`ınh: y2 y0 = − 2 (1) x2 0 0 D- ˘a.t y = ux ⇒ y = u x + u, ta c´o: (1)⇔ u0x + u = u2 − 2 ⇔ u0x = u2 − u − 2 (2)
- 43 u = −1 y = −x +Nˆe´u u2 − u − 2 ≡ 0 ⇔ th`ı u0 = 0, (2) tho’a m˜an,vˆay l`ac´ac u =2 . y =2x nghiˆe.mcu’a (1) u =6 −1 +Nˆe´u u2 − u − 2 6≡ 0 ⇔ th`ı(2) tu.o.ng d¯u.o.ng v´o.i: u =26 du dx 1 u − 2 1 u − 2 = ⇒ ln + ln C ⇔ ln = Cx3 u2 − u − 2 x 3 u − 1 3 u +1 " y − 2x = Cx3(y + x) ⇔ y−2x = Cx3(y+x), ∀C =6 0. Suy ra c´acnghiˆe.mcu’a (1) l`a: y = −x v´o.i C t `u y ´y . 2.3. Phu.o.ng tr`ınhvi phˆantuyˆe´n t´ınhcˆa´p1 . . 0 L`aphuong tr`ınh c´oda.ng y + p(x)y = q(x) trong d¯´o p(x),q(x) l`ac´ach`amliˆen tu. c trˆen [a, b]. . . . C´ach gia’i thu. chiˆe.n qua c´acbu´oc: . . − Gia’iphuong tr`ınhtuyˆe´n t´ınhthuˆa`n nhˆa´t(q(x) = 0), ta c´o: y ≡ 0 ho˘a.c dy = −p(x)dx ⇒ y = Ce− R p(x)dx,vˆay nghiˆem l`a: y = Ce− R p(x)dx y . . ∗ . . − T`ım nghiˆe.m riˆeng y cu’aphuong tr`ınhkhˆongthuˆa`n nhˆa´t(q(x) =6 0) b˘a`ng ∗ . − p(x)dx c´ach d¯˘a.t y = C(x).u(x)v´oi u(x)=e R , suy ra y∗ = e− R p(x)dx. Z q(x)eR p(x)dxdx. . . ∗ − lˆa.p nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınh khˆongthuˆa`n nhˆa´tda.ng y = y + y . . V´ıdu. 6. Gia’iphuong tr`ınh: y0 − 2xy = x . . 0 + Gia’iphuong tr`ınhthuˆa`n nhˆa´t y − 2xy = 0, ta c´onghiˆe.m: dy 2 y = 0 ho˘ac =2xdx ⇒ ln y = x2 +lnC ⇒ y = Cex . . y . . + Nghiˆe.mriˆeng cu’aphuong tr`ınhkhˆongthuˆa`n nhˆa´t l`a: 2 2 2 1 2 1 y∗ = ex . Z x.e−x dx = ex . − e−x = − . 2 2 2 1 Vˆay (1) c´onghiˆemtˆo’ng qu´at: y = y + y∗ = Cex − v´o.i C t `u y ´y . . . 2 . . V´ıdu. 7. Gia’iphuong tr`ınh: 2 y0 +2xy = xe−x . . . 0 + Gia’iphuong tr`ınhthuˆa`n nhˆa´t y +2xy = 0, ta c´onghiˆe.m: dy 2 y = 0 ho˘ac = −2xdx ⇒ ln y = −x2 +lnC ⇒ y = Ce−x . . y
- 44 . . + Nghiˆe.mriˆeng cu’aphuong tr`ınhkhˆongthuˆa`n nhˆa´t l`a: 2 −x2 2 2 2 2 x e y∗ = e−x . Z x.e−x .ex dx = e−x . Z xdx = . 2 2 −x2 2 x e Vˆay (1) c´onghiˆemtˆo’ng qu´atl`a: y = y + y∗ = Ce−x + v´o.i C t `u y ´y . . . 2 2.4. Phu.o.ng tr`ınhBernoulli . . 0 α L`aphuong tr`ınhc´oda.ng y + p(x)y = q(x).y . y1−α D- ˆe ’ gia’i, gia’ thiˆe´t y 6≡ 0, chia 2 vˆe´ cho yα,rˆo`id¯˘at z = (l`ah`amtheo x, z 6≡ 0), . 1 − α gia’iphu.o.ng tr`ınhtuyˆe´n t´ınhcˆa´p 1 theo z. . . V´ıdu. 8. Gia’iphuong tr`ınh: y0 +2xy =2x3y3. 0 . . +Nˆe´u y ≡ 0th`ıy = 0: (1) tho’a m˜annˆen y = 0 l`anghiˆe.mcu’aphuong tr`ınh 1 +Nˆe´u y 6≡ 0 (1)⇒ y0y−3 +2xy−2 =2x3.D- ˘at z = − y−2 (l`ah`amtheo x, z 6≡ 0), . 2 th`ı: z0 = y0y−3,phu.o.ng tr`ınh tro’. th`anh z0 − 4xz =2x3 (3) Gia’i (3). Phu.o.ng tr`ınhthuˆa`n nhˆa´t: dz 2 z0 − 4xz =0⇒ =4xdx ⇒ z = Ce2x z v`anghiˆe.mriˆeng 2 2 2 1 1 2 1 1 z∗ = e2x Z 2x3e−2x dx = e2x − x2 + e−2x = x2 + . 2 2 2 2 2 1 1 Vˆay (3) c´onghiˆemtˆo’ng qu´at: z = z +z∗ = Ce2x − x2 + , nˆen(1) c´onghiˆem . . 2 2 . 1 2 1 = −2Ce2x + x2 + y2 2 ,v´o.i C t `u y ´y . y =0 BAI` TAˆ. P . . . 4.1.1. Gia’i c´acphuong tr`ınhvi phˆansau (da. ng d¯uavˆe` biˆe´nsˆo´ phˆanly): x + y x − y (xy2 −x)dx+(y+x2 y)dy =0; y0 +sin −sin =0; y0 =2x+y+4; 2 2 √ 2x2 5y y0 = y − x +1; y0 = ex+y−1; xy0 = ey −1; dx+ dy =0; 1+2x2 y2 +1 (1 + e2x)y2dy = exdx (biˆe´t y(0) = 0); y0 = ey−4x (biˆe´t y(1) = 1) . . 4.1.2. Gia’i c´acphuong tr`ınhvi phˆansau (da. ng d¯˘a’ ng cˆa´pcˆa´p 1): 2 0 y 0 y y 0 y 0 2xy y = − 2; y = e x + ; xy = y ln ; y = ; x2 x x x2 − y2 √ y y y x (x2+2xy)dx+xydy =0; xy0 = y− xy; y0 = 1+ln ; y0 = + ; x x x y
- 45 y y y y π y0 = + cos2 ; x3y0 = y(x2 + y2); y0 = + sin (biˆe´t y(1) = ); x x x x 2 y π y y2 xy0 − y = xtg (biˆe´t y(1) = ); y0 = + (biˆe´t y(−1) = 1); x 2 x x2 y 1 y 3 y0 = + (biˆe´t y(−1) = 1); x 2 x . . 4.1.3. Gia’i c´acphuong tr`ınhvi phˆansau (da. ng tuyˆe´n t´ınhcˆa´p 1): y y0 +2y =4x; (1 + x2)y0 − 2xy =(1+x2)2; xy0 − = x; 1+x 2 xy0 + y = x2 cos x; y0 +2xy = xe−x ; y0 cos x + y sin x =1; 1 2 xy0−xy = (1+x2)ex; y0+exy = e2x; y0− y = x ln x; y0− y =4x2; x ln x x y y0 + xy =3x; y0 + =3x3; y0 +2y = cos x; y0 − 2y = sin x; x xy0 + y = ex (biˆe´t y(1) = 0); (x +1)xy0 − y = x(x +1) (biˆe´t y(1) = 0) II. Phu.o.ng tr`ınhvi phˆancˆa´p2 1. Kh´ainiˆe.m chung . . . . *Tago.i phuo ng tr`ınh vi phˆancˆa´p2l`aphuong tr`ınhc´oda.ng F (y00,y0,y,x) = 0 (II) hay 00 0 y = f(y ,y,x)(IIo) 0 00 trong d¯´o y l`ah`amsˆo´ theo biˆe´n x, c`on y ,y l`ad¯a.o h`amcˆa´p 1,2 cu’a y,v`anghiˆe.m cu’a phu.o.ng tr`ınh l`ah`am y = ψ(x) hay Φ(x, y) = 0 tho’a m˜anphu.o.ng tr`ınhd¯´o. . . * H`am y = ψ(x, C1,C2) ho˘a.cΦ(x, y, C1,C2)=0tho’a m˜anphuong tr`ınh(II)hay . . (IIo)v´oi C1,C2 l`ah˘a`ng sˆo´ t`uy´ytrong tˆa.p con n`aod¯´ocu’a R, v`av´oimˆo˜id¯iˆ`eu 0 0 . . kiˆe.n y(xo)=yo v`a y (xo)=yo ta t`ımd¯uo. c duy nhˆa´tc˘a.psˆo´ C10,C20 sao cho y = . . ψ(x, C10,C20) hay Φ(x, y, C10,C20) = 0 tho’a(II)hay(IIo)d¯uo. cgo.il`anghiˆe.m tˆo’ng qu´atcu’a c´acphu.o.ng tr`ınh d¯´o. *Nˆe´u y = ψ(y,C1,C2) hay Φ(x, y, C1,C2) = 0 l`anghiˆe.mtˆo’ng qu´atcu’a(II)hay . (IIo), cho C1 = C01,C2 = C02 v´oi C01,C02 l`ahai sˆo´ x´acd¯i.nh cu. thˆe’ th`ı y = . . . . ψ(x, C01,C02) hay Φ(x, y, C01,C02)=0d¯uo. cgo.il`anghiˆe.m riˆengcu’a phuo ng tr`ınh d¯´o. . . + (D- .inh l´ytˆo`nta.i v`aduy nhˆa´t nghiˆe.m): Trong phuong tr`ınh(IIo), nˆe´u h`am 0 . ’ 0 f(y ,y,x) liˆentu.c trong miˆe` n n`aod¯´och´uad¯iˆem (yo,yo,xo) th`ıtˆo`nta.imˆo.t nghiˆe.m 0 0 0 0 y = y(x) cu’a(IIo) sao cho y + o = y(xo),yo = y (xo) v`anˆe´u fy.fy0 c˜ung liˆentu.c . ’ 0 trong miˆ`ench´uad¯iˆem (yo,yo,xo) th`ınghiˆe.mˆa´y l`aduy nhˆa´t. . . . . 2. C´acloa.i phuo ng tr`ınh vi phˆancˆa´p 2 thu`o ng g˘a.p . . . . 2.1. Phuong tr`ınhvi phˆancˆa´p2gia’mcˆa´pd¯uo. c . . 00 0 + Phuong tr`ınhc´oda.ng y = f(x) (thiˆe´u y,y ) C´ach gia’i: t´ıch phˆan2 lˆa`n. . . V´ıdu. 1. Gia’iphuong tr`ınh: y00 = x +1.
- 46 x2 Ta c´o: y0 = Z (x +1)dx = + x + C , suy ra: 2 1 x2 x3 x2 y = Z + x + C dx = + + C x + C v´o.i C ,C t `u y ´y . 2 1 6 2 1 2 1 2 . . 00 0 + Phuong tr`ınhc´oda.ng y = f(y ,x) (thiˆe´u y) 0 00 0 0 . . C´ach gia’i: d¯˘a.t y = z (h`amtheo x) ⇒ y = z .Nˆen: z = f(z,x) l`aphuong 0 tr`ınh cˆa´p1cu’a z theo x, gia’i ra nghiˆe.mtˆo’ng qu´at z = ψ(x, C1), thay z = y , ta c´o: 0 . . y = ψ(x, C1) gia’i ra nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhban d¯ˆa`u. . . V´ıdu. 2. Gia’iphuong tr`ınh: y00 = y0 + x. 0 00 0 0 . . D- ˘a.t y = z (h`amtheo x) ⇒ y = z , suy ra z − z = x.D- ˆayl`aphuong tr`ınhvi . phˆantuyˆe´n t´ınhcˆa´p1cu’a h`am z theo x v´oi p(x)=−1,q(x)=x nˆenc´onghiˆe.m: Z R p(x)dx − R p(x)dx Z −x x x z = q(x)e dx + C1 e = xe dx + C1 e = C1e − (x + 1). x2 Thay z = y0, ta c´o: y0 = C ex − (x +1)⇒ y = C ex − − x + C v´o.i C ,C t`uy´y. 1 1 2 2 1 2 . . 00 0 + Phuong tr`ınhc´oda.ng y = f(y,y ) (thiˆe´u x) 0 00 0 0 0 C´ach gia’i: d¯˘a.t y = z (h`amtheo y), d¯a.o h`amtheo x, ta c´o: y = zy · y = z · z, nˆen: z0 · z = f(y,z). . . 0 Gia’iphuong tr`ınh cˆa´p1cu’a z theo biˆe´n y, ta c´o: z = ψ(y,C1), thay z = y rˆo`i gia’i . . 0 . . tiˆe´pphuong tr`ınh y = ψ(y,C1) ta c´onghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhd¯˜acho. . . V´ıdu. 3. Gia’iphuong tr`ınh: (1 − y)y00 +2(y0)2 = 0 (1) dz D- ˘at y0 = z (theo y) ⇒ y00 = z0y0 = z0z (v´o.i z0 = ), ta c´o: . dy (1 − y)z0z +2z2 = 0 (2) 0 0 +Nˆe´u z ≡ 0 ⇒ z = 0: (2) tho’a m˜annˆen z ≡ 0 l`anghiˆe.mcu’a (2)⇒ y =0⇒ y = C1 . (v´oi C1 t`uy´y)l`anghiˆe.mcu’a (1) 0 . . +Nˆe´u1− y ≡ 0 ⇔ y ≡ 1 ⇒ y = 0: (1) tho’a m˜annˆen y = 1 l`anghiˆe.m (1) (tru`ong . ho.priˆeng cu’a nghiˆe.m y = C1) +Nˆe´u y 6≡ C1 ⇔ z 6≡ 0: dz dz 2dy (2)⇒ (1 − y) = −2z ⇒ = ⇒ ln |z| =2ln|y − 1| +ln|C | dy z y − 1 1 dy 1 Suy ra: z = y0 = C (y − 1)2 ⇒ = C dx ⇒− = C x + C . 1 (y − 1)2 1 y − 1 1 2 1 y = − +1;C1 =06 ,C2 t `u y ´y Vˆa.y (1) c´onghiˆe.m: C1x + C2 y = C1,C1 t `u y ´y . . . 2.2. Phuo ng tr`ınh vi phˆantuyˆe´nt´ınh cˆa´p2v´oihˆe. sˆo´ h˘a`ng
- 47 . . 00 0 . L`aphuong tr`ınh c´oda.ng y + py + qy = f(x) trong d¯´o p, q l`ah˘a`ng sˆo´ thu. c. * D- ˆo´i v´o.i phu.o.ng tr`ınhthuˆa` n nhˆa´t(f(x)=0): . . . 2 Gia’iphuong tr`ınhd¯˘a.c trung: k + pk + q = 0. (DT) . . . +Nˆe´u (DT) c´o2 nghiˆe.m thu. c phˆanbiˆe.t k1,k2 th`ınghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınh thuˆa`n nhˆa´t l`a: k1x k2x y = C1e + C2e . . . +Nˆe´u (DT) c´onghiˆe.mk´ep k1 = k2 th`ınghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhthuˆa`n nhˆa´t l`a: k1x y =(C1 + C2x)e . . +Nˆe´u (DT) c´o2 nghiˆe.mph´uc k1 = α + βi,k2 = α − βi th`ınghiˆe.mtˆo’ng qu´atcu’a phu.o.ng tr`ınh thuˆa`n nhˆa´t l`a: αx y = e (C1 cos βx + C2 sin βx). . . . 00 0 * D- ˆo´iv´oi phuong tr`ınhkhˆongthuˆa` n nhˆa´t y + py + qy = f(x) (vˆe´ pha’i c´oda.ng d¯˘a.c biˆe.t): . . . . . . . . . Bu´o c1: Gia’iphuong tr`ınh thuˆa`n nhˆa´ttuong ´ung, t`ımnghiˆe.mtˆo’ng qu´atdu´o ida.ng: y = C1y1(x)+C2y2(x) . . ∗ . . Bu´o c2:T`ım nghiˆe.mriˆeng y cu’aphuong tr`ınhkhˆongthuˆa`n nhˆa´td¯ˆe’ suy ra nghiˆe.m y = y + y∗ ax . +Nˆe´u f(x) c´oda.ng Pn(x)e (Pn(x) l`ad¯ath´ucbˆa.c n): ∗ − Nˆe´u a khˆongpha’i l`anghiˆe.mcu’a (DT) th`ı y c´oda.ng: ∗ n n−1 ax y =(anx + an−1x + ···+ a1x + ao)e . ∗ − Nˆe´u a l`anghiˆe.md¯oncu’a (DT) th`ı y c´oda.ng: ∗ n n−1 ax y = x(anx + an−1x + ···+ a1x + ao)e ∗ − Nˆe´u a l`anghiˆe.mk´epcu’a (DT) th`ı y c´oda.ng: ∗ 2 n n−1 ax y = x (anx + an−1x + ···+ a1x + ao)e ax . +Nˆe´u f(x) c´oda.ng e [Pn(x) cos bx + Qm(x) sin bx]: (Pn(x),Qm(x) l`ac´acd¯ath´uc bˆa.c n, m), d¯˘a.t h = max{m, n}: ∗ − Nˆe´u a + bi khˆongpha’i l`anghiˆe.mcu’a (DT) th`ı y c´oda.ng: ∗ h h ax y = (ahx + ···+ a1x + ao) cos bx +(bhx + ···+ b1x + bo) sin bx e ∗ − Nˆe´u a + bi l`anghiˆe.mcu’a (DT) th`ı y c´oda.ng: ∗ h h ax y = x. (ahx + ···+ a1x + ao) cos bx +(bhx + ···+ b1x + bo) sin bx e . . . ∗0 ∗00 D- ˆe ’ x´acd¯i.nh c´acsˆo´ ai,bi o’ trˆen, ta d`ungphuong ph´aphˆe. sˆo´ bˆa´td¯i.nh: t´ınh y ,y ∗ ∗0 ∗00 . . rˆo`i thay y ,y ,y v`aophuong tr`ınhkhˆongthuˆa`n nhˆa´t, d¯ˆo`ng nhˆa´t hai vˆe´ v`agia’ihˆe. . . phuong tr`ınhtheo ai,bi.
- 48 . . + Nguyˆenl´y chˆo`ng chˆa´t nghiˆe.m: Nˆe´u y1(x),y2(x) lˆa` nluo. t l`anghiˆe.m riˆengcu’a . . 00 0 00 0 c´acphuong tr`ınh y + p(x).y + q(x).y = f1(x) v`a y + p(x).y + q(x).y = f2(x) th`ı 00 0 y1(x)+y2(x) l`anghiˆe.m riˆengcu’a y + p(x).y + q(x).y = f1(x)+f2(x). . . V´ıdu. 1. Gia’iphuong tr`ınh: y00 − 2y0 − 3y = e4x (1) k = −1 . . . 2 1 . . Phuong tr`ınhd¯˘a.c trung k − 2k − 3 = 0 c´onghiˆe.m nˆen phuong tr`ınh k2 =3 00 0 −x 3x thuˆa`n nhˆa´t: y − 2y − 3y = 0 c´onghiˆe.m y = C1e + C2e , C1,C2 t`uy´y. ax . Vˆe´ pha’i (1) c´oda.ng Pn(x)e v´oi n =0,a =4=6 k1,k2 nˆennghiˆe.mriˆeng c´oda.ng ∗0 4x ∗ 4x y =4aoe y = aoe , suy ra: ∗00 4x . Thay v`ao(1), ta c´o: y =16aoe 1 16a e4x − 8a e4x − 3a e4x = e4x ⇒ a = , suy ra: o o o o 5 1 y = y + y∗ = C e−x + C e3x + e4x, ∀C ,C . 1 2 5 1 2 . . V´ıdu. 2. Gia’iphuong tr`ınh: y00 − 2y0 + y =6xex (2) . . . 2 . . Phuong tr`ınhd¯˘a.c trung k − 2k + 1 = 0 c´onghiˆe.mk´epk1 = k2 = 1 nˆenphuong 00 0 x tr`ınh thuˆa`n nhˆa´t: y − 2y + y = 0 c´onghiˆe.m y =(C1x + C2)e , C1,C2 t `u y ´y . ax . Vˆe´ pha’i (2) c´oda.ng Pn(x)e v´oi n =1,a =1=k1 = k2 nˆennghiˆe.mriˆeng c´oda.ng ∗0 3 2 x ∗ 2 x y =[a1x +(3a1 + ao)x +2aox]e y = x (a1x + ao)e , suy ra: ∗00 3 2 x . y =[a1x +(6a1 + ao)x +(6a1 +4ao)x +2ao]e 6a1 =6 a1 =1 ∗ Thay v`ao(2), ta c´o: ⇒ ⇒ y = x3ex nˆen(2) c´onghiˆe.m: 2ao =0 ao =0 ∗ x 3 x 3 x y = y + y =(C1x + C2)e + x e =(C1x + C2 + x )e , ∀C1,C2 . . V´ıdu. 3. Gia’i p[huong tr`ınh: y00 + y =4xex (3) . . . 2 . . Phuong tr`ınhd¯˘a.c trung k + 1 = 0 c´onghiˆe.m k = ±i nˆen phuong tr`ınhthuˆa`n 00 0x nhˆa´t: y + y = 0 c´onghiˆe.m y = e (C1 sin x + C2 cos x)=C1 sin x + C2 cos x, C1,C2 t `u y ´y . ax . Vˆe´ pha’i (3) c´oda.ng Pn(x)e v´oi n =1,a=1=6 k1,k2 nˆennghiˆe.mriˆeng c´oda.ng: ∗0 x ∗ x y =(a1x + a1 + ao)e y =(a1x + ao)e , suy ra: ∗00 x . y =(a1x +2a1 + ao)e Thay v`ao(3), ta c´o:2a1x +2a1 +2ao =4x.D- `ˆong nhˆa´t2vˆe´: a1 =2 a1 =2 ∗ ⇒ ⇒ y =(2x − 2)ex nˆen(2) c´onghiˆe.m: a1 + ao =0 ao = −2 ∗ x y = y + y =(C1 sin x + C2 cos x)+(2x − 2)e , ∀C1,C2 . . V´ıdu. 4. Gia’iphuong tr`ınh: y00 − y =2ex − x2 (4) . . . 2 . . Phuong tr`ınhd¯˘a.c trung k − 1 = 0 c´onghiˆe.m k = ±1 nˆen phuong tr`ınhthuˆa`n 00 x −x nhˆa´t: y − y = 0 c´onghiˆe.m y = C1e + C2e , C1,C2 t `u y ´y .
- 49 Theo nguyˆenl´ychˆo`ng chˆa´t nghiˆe.m, nghiˆe.mriˆeng cu’a (4) l`atˆo’ng hai nghiˆe.mriˆeng y00 − y =2ex (4a) cu’a hai phu.o.ng tr`ınhsau: y00 − y = −x2 (4b) ax . Vˆe´ pha’i (4a) c´oda.ng Pn(x)e v´oi n =0,a =1=k1 nˆennghiˆe.mriˆeng c´oda.ng: ∗0 x ∗ x x y1 =(aox + ao)e y1 = x(ao)e = aoxe , suy ra: ∗00 x . y1 =(aox +2ao)e x x ∗ x Thay v`ao(4a), ta c´o:(aox +2ao − aox)e =2e ⇒ ao = 1, nˆen y1 = xe ax . Vˆe´ pha’i (4b) c´oda.ng Pn(x)e v´oi n =2,a=0=6 k1,k2 nˆennghiˆe.mriˆeng c´oda.ng: a2 =1 y∗0 =2a x + a ∗ 2 2 2 1 (4⇒b) ⇒ ∗ 2 y2 = a2x + a1x + ao, suy ra: ∗00 a1 =0 y2 = x +2. y2 =2a2 ao =2 ∗ ∗ ∗ x 2 ’ Suy ra nghiˆe.mriˆeng cu’a (4) l`a: y = y1 + y2 = xe + x + 2 v`anghiˆe.mtˆong qu´at: ∗ x −x x 2 y = y + y = C1e + C2e + xe + x +2, ∀C1,C2 BAI` TAˆ. P . . 4.2.1. Gia’i c´acphuong tr`ınhvi phˆancˆa´p 2 sau (da.ng gia’mcˆa´p): y0 xy00 = y0; xy00 = y0 ln ; x2y00 = y02; y3y00 =1; y00(ex +1)+y0 =0; x (x ln x)y00 − y0 =0; x2y00 +3xy0 =0; 1+y02 =2yy00; yy00 − y02 =0 . . . 4.2.2. Gia’i c´acphuong tr`ınhvi phˆancˆa´p 2 sau (da.ng tuyˆe´n t´ınhv´oihˆe. sˆo´ h˘a`ng): y00 −2y0 +y = ex; y00 −5y0 +6y = e2x; y00 −2y0 +2y =2x2; y00 +y0 −2y = xex; y00 − 3y0 +2y = ex(2x +3); y00 − y0 − x; y00 − 6y0 +5y =3ex +5x2; y00 − 5y0 =3x2 + sin 5x; y00 + y = sin x cos 3x; y00 − 2y0 − 3y =3− 4ex -ooOoo-
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