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Các phương pháp giải nhanh đề thi đại học

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  1. Toặn hổc phưí thưng a
  2. ToanDHSP.COM Li nĩi đu: o oo oo o oo o ơ o • ơ • ơ o ơo o o o o ơo ơ o • ơ o oo o oo oooo oơ ơơo ơoo oo o o oo o a ayya yyaye e
  3. Bài I: ơ ơ ư ươ  ayey n yy  y  a n ey y y ươa  aye a aa x= x0 + a 1 t  y= y0 + a 2 t  a a x=3 + 2 t  y=4 + 3 t yaa  ey n  e a a x=2 + t   y=2 − t x3 +8 + 3 12 −x3 = 10 x3 + 8 12 − x3  ya  aay
  4. ToanDHSP.COM y ayaay 3 3 x+8 + 312−x = 10 X Y a aea X = 1+ 3t  Y = 3 - t yyy yy 3 x+3 x+2 X Y  x + 3 = 1− t x + 3 = 1− 2t + t 2    3 3 x + 2 = t x + 2 = t ya  •  ư y y  x + 3 = u • a  y e 3 x + 2 = v yya • ayaa aa a x + y − xy = 3 (1)   x +1 + y +1 = 4 (2)  x +1 = 2 + t x +1 = t 2 + 4t + 4    2  y +1 = 2 − t y +1 = t − 4t + 4 x = t 2 + 4t + 3   y = t 2 − 4t + 3 (t 2 + 3 + 4t)(t 2 + 3 − 4t)
  5.  t4 −10 t 2 + 9    hoặc  y a f() x= m a  x+2 m = 1 + 3 t   3m− x = 3 − t x+2 m = 1 + 6 t + 9 t 2    3m− x = 9 − 6 t + t 2   y 1) 2)  2x+ y + 1 − x + 1 = 1 3)  3x+ 2 y = 4 4) 1− sin(x ) + 1 + cos( x ) = 1
  6. ToanDHSP.COM Bài II: Các cách giải phương trình và bất phương trình vơ tỉ. ya ya aya yy ayeay a • • yaa • B ≥ 0  A = B   A = B 2 B ≥ 0  A B  B ≥ 0   2 A > B VD1. x ≥ 0  5 − x ≥ 0 0 ≤ x ≤ 5 0 ≤ x ≤ 5       2 2 10 − x ≥ 0 2 5x − x 2 = 5 − x 4(5x − x ) = 25 −10x + x  x + 5 − x + 2 x 5(− x) = 10 0 ≤ x ≤ 5    ∨ x 2 − 6x + 5 = 0 2 x − x + 3 x −1 x ≥ 1 x ≥ 1      x 2 + 2x − 3 > x 2 − 2x +1 x > 1
  7. VD3.       VD4. −x2 + x = 0   x− x 2 > 0 ⇔ x= 0∨ x = 1   2 x− x −2 ≥ 0 ư yaya aayaaa a ea B = 0  B  B > 0 a  A ≥ 0 VD5.     3x − 5  2 −   ≥ 0   4    3x − 5 ≥ 0  x=3   3x − 5  2 x 2 −8 =     4 
  8. ToanDHSP.COM ư aay yaayya yayay aaaa aayaaa ươ  f (u(x); n u(x))≥ 0 f (u(x); n u(x))≤ 0 n u(x)  f (u(x); n u(x)) = 0  VD1. 2 2  => t>0 ; t +2= x + x    VD2. t ≥ 0 x −1   t 2 +1 = x   t2+1-(t+1)=2  t2-t-2=0  t=2 hoặc t=-1 x=5 VD3.  =>
  9. ∨ t=-3   TH2: t=-3    f(n u()() x+ n v x ) n u() x= u  a m v() x= v 23 3x − 2 + 3 6 − 5x − 8 = 0 y 5 8 3 3x− 2 = u  u3+ v 2 =   3 3  6− 5x = v ( v ≥ 0)  2u+ 3 v − 8 = 0  2 5 3 2 8 5 3  8− 2u  8 2  u+ v =  u +   = (u+ 2)(15 u − 26 u + 20) = 0 3 3 3  3  3        8− 2u  8− 2u  8− 2u v = v = v =  3  3  3 u = −2    v = 4 A yaaya y ayy
  10. ToanDHSP.COM  11 x − = y − (1)  x y  3 2y = x +1 (2) y  1  x = y a (1) ⇔ ( x − y)1+  = 0 ⇔   xy  xy = −1  xy== 1  xy=x=y xy=  −+1 5  ⇔⇔⇔== xy 2yx=xx 33+1 2 = +1 xx−x+1− 2 1 = 0  2   ( )( )   −1− 5 xy= =  2  1  1  y =− 2 2 xy =− 1  x y =− 4  2 1   1  3 x x x x x ⇒ VN  3 ⇔⇔ x + + 2 =  −  +  +  + > 0, ∀ 21yx=+  2 3 4  2   2  2 −=+ x 1 xx + + 2 = 0  x  −1+ 5 −1+ 5   −1− 5 −1− 5  ya ( x; y) = (1;1), ; ,  ;   1 1   1 1  2 x +1+ y(y + x) = 4y (1)  x, y∈R . 2 ( ) (x +1)(y + x − 2) = y (2) (1) ⇔ x2 +1+ y ( x + y − 4) = 0 (*) u = x2 +1 > 0; v = x + y − 4 u − yv = 0 (3) ⇔  aya (3) ⇔ u + u (v + 2).v = 0 ⇔ u 1+ v(v + 2) = 0 u (v + 2) = y (4) ⇔ v2 + 2v +1 = 0 ⇔ (v +1)2 = 0 ⇔ v = −1 ⇔ x + y = 3 2 ⇒ xy +−1= 0 2 xy=1 =− 2 y ⇔⇔ +−−=⇔ xx 1 (3 ) 0  xy= 3− xy= 2 ⇒ = 5 x3 −8x = y3 + 2y  x, y∈ R . 2 2 ( ) x − 3 = 3(y +1) (*) 3 3 x3 − y3 = 2(4x + y) 3( x − y ) = 6(4x + 2y) (1) ⇔  ⇔  yaya 2 2 2 2 x − 3y = 6 x − 3y = 6 (2) ⇔ 3( x3 − y3 ) = ( x2 − 3y2 )(4x + y) ⇔ x3 −12y2 x + x2 y = 0 ⇔ x( x2 + xy −12y2 ) = 0 yyayayya
  11. x2 + xy −12 y2 = 0  ( x−3 y)( x + 4 y) = 0 ⇒ ⇔  x2−3 y 2 = 6 x2−3 y 2 = 6 x−3 y = 0  x = 3 y  y= 1⇒ x = 3 ⇔  ⇔  x2−3 y 2 = 6  6 y2 = 6  y= −1⇒ x = − 3  78− 4 78  y= ⇒ x = x= −4 y x= −4 y 13 13  ⇔ ⇔  x2−3 y 2 = 6 13 y2 = 6  78 4 78  y= − ⇒ x =  13 13 ya  78− 4 78  − 78 4 78  ()()()x; y = 1;3,1;3, − −  ; ,  ;   13 13  13 13   2 2 ()x− y( x + y ) = 13() 1  2 2 ()x+ y() x − y = 25() 2 aaay 2 2 2 2 2 2  ⇔+13(xyxy )()() −−− 25 xyxy( +=⇔−) 0()() xy 13 xy +− 25( xy +=) 0 ⇔−−+( x y)( 12 x2 26 xy − 12 y2 ) =⇔−−−+ 0 2( x y)( 12 x2 26 xy − 12 y2 ) = 0 yya 3x= 2 y  y = −3 25 2 −y  ⇔  3x= 2 y  y .  = 25 x = −2 ()()3x− 2 y 2 x − 3 y = 0   9 3  ⇒ ⇔ 2x= 3 y ⇔  2 2 2x= 3 y ()x+ y() x − y = 25  2  ()()x+ y x − y = 25   x = 3 252  1   y. y  = 25 ⇔   4 2  y = 2 aaa(−12x2 + 26 xy − 12 y2 ) (3x− 2 y)( 2 x − 3 y) yaayya ayyyae (−12x2 + 26 x − 12) = 0 y y a 3 2 x= ∨ x = y a y y a 2 3 3 2 x= y ∨ x = y ya 2 3 (−12x2 + 26 xy − 12 y2 ) = 0 ⇔ (3x− 2 y)( 2 x − 3 y) = 0 y (−12x2 + 26 xy − 12 y2 ) = − 23223( x − y)( x − y) = 0  aaa aaa aae
  12. ToanDHSP.COM ayya aey y x43−22 x y + x y =1 x43+2 2x y + x y2 = 2x + 9 Bài 1.  Bài 7.  x32y −+x= xy 1 x2 +=2+xy 6x 6 x2 + y2 + x + y = 4 xy + x +1 = 7y Bài 2.  Bài 8.   2 2 2 x( x + y +1) + y ( y +1) = 2 x y + xy +1 =13y 3 x2 −+x=y− y2 3( x y)  x +1 − y = 8 − x  Bài 9. Bài 3.   4 2 2 2 x ++x=y y 7( x − y) ( x −=1) y 2 log xx3 x+y2 2 − 3 −5 = 3  y + 2  x ( ) 3y = Bài 4.   x2 log yy3 y+ 2 2 − 3 − 5x = 3 Bài 10.   y ( ) x2 + 2 3x = x( x + y +1) − 3 = 0  y2 Bài 5.   2 5  x y 1 1 ( + ) − 2 +1 = 0 x − = y −  x Bài 11.  x y 99  3 xy += 1 2y =+x 1 Bài 6.   x25 +=y+25 x16 y16
  13. Bài III: ơ ư 2 2 2 1 2 1 1. sin x+ cos x = 1;1 + tanx = ;1 + cotx = . cos2 x sin 2 x sinx cosx 1 2. tanx= ;cot x = ;tan x = . cosx sin x cot x sin(a± b ) = sin a cos b ± cos asinb cos(a± b ) = cos a cos b∓ sin a sin b o ooo 2 1+ cos2x 2 1− cos2x cos x = ;sin x = 2 2 ooo o 2tan x 1− tan 2 x 2tan x sin 2x =;cos2x =;tan 2x = 1+ tan 2x 1 + tan 2x 1 − tan 2 x 1 cosa cos b=( cos( a − b ) + cos( a + b )) 2 1 sina sin b=( cos( a − b ) − cos( a + b )) 2 1 sina cos b=( sin( a − b ) + sin( a + b )) 2 x+ y x − y sinx+ sin y = 2sin cos 2 2 x+ y x − y sinx− sin y = 2cos sin 2 2 x+ y x − y cosx+ cos y = 2cos cos 2 2 x+ y x − y cosx− cos y = − 2sin sin 2 2
  14. ToanDHSP.COM ươư a yaya ye yya  π  2sin  2xx−  + 4sin +1 = 0  6   3 sin 2x − cos 2x + 4sin x +1 = 0  2sin x( 3 cos 2x + 2) − 2sin2 x = 0 sinx = 0 ⇔ x = kπ   2sin x( 3 cos x − sin x + 2) = 0  1  π   3 cos x − sin x = −1 ⇔ cos x +  = cos x  2  6   xk= π  5π  x =+ 2kπ  6  −7π x =+ 2kπ  6 π a x 3π 4sin 2 − 3 cos 2x = 1+ 2cos2 (x − ) 2 4 ∈(0,π) 2 x 2  3π  a 4sin − 3 cos2x = 1+ 2 cos  x −  2  4   3π  ⇔ 2(1− cosx) − 3 cos2x = 1+1+ cos2x −   2  ⇔ 2 − 2cosx − 3 cos2x = 2 − sin2x ⇔ −2cosx = 3 cos2x − sin2x aa 3 1 ⇔ − cosx = cos2x − sin 2x 2 2  π 5π 2π 7π ⇔ cos2x +  = cos(π − x) ⇔ x = + k (a) hay x = − + h2π (b)  6  18 3 6
  15. x∈( 0, π) a a a 5π 17 π 5 π (0,π) x= ,x = ,x = 118 2 183 6 π 2 2 cos3 (x − ) − 3cosx − sin x = 0 4 3   π  ⇔2cosx −   − 3cosx − sinx = 0 4   ⇔cosx + sinx3 − 3cosx − sinx = 0 () ⇔cos3 x + sin 3 x + 3cos2 xsinx+ 3cosxsin2 x− 3cosx − sinx = 0 cosx= 0 cosx≠ 0 ⇔  hay sin3 x− sinx = 0 1+ 3tgx + 3tgx2 + tgx 3 − 3 − 3tgx 2 − tgx − tgx 3 = 0 2 π π ⇔sin x = 1 haytgx= 1 ⇔x = + k π ay x= + k π 2 4 π cos 2x − 1 tg(+ x ) − 3 tg2 x = 2 cos2 x −2sin2 x ⇔ −cot gx − 3tg2 x = cos2 x 1 π ⇔− −tgx02 =⇔ tgx3 =−⇔ 1 tgx =−⇔=−+π∈ 1 x k,kZ tgx 4 ƯOƯƯ ∈ sin2 x t 2 a cos2 x 1− t 2 1− 2sin2 x 3sinx− 4sin3 x = 3 t − 4 t3 sin2 x= 1 − cos2 x = 1 − t 2 cos 2x= 2 t 2 + 1 sin2 x 1− t 2 tan2 x = = cos3x= 4cos3 x − 3cos x = 4 t 3 − 3 t cos2x t 2 a
  16. ToanDHSP.COM 1 1 cot x = cos2 x = t 1+ t 2 t 2 1− t 2 sin2 x = cos 2x = 1+ t 2 1+ t 2  1  2t sin2x=2t   t an2x = 1+ t 2  1+ t 2 asin x + b cos x a tan x + b at + b = = csin x + d cos x c tan x + d ct + d   ∈ − 2; 2 t 2 −1 = ± t 2 +1 ±2 ( )  t 2 −1 3− t3 sin3 x + cos3 x = (sin x + cos x)(sin2 x + cos2 x − sin x cos x) = t 1−  =  2  2 ƯƯ   y  ya     cos21x cot x −1 = + sin2 x − sin 2x 1+ tan x 2 a 1− t 2    112 1+ t 2  tt2 −1 = + − (t ≠ 0;t ≠ −1) t 1+ t 1+ t 22 2 1+ t π ⇔ 2t3 − 3t 2 + 2t −1 = 0 ⇔ t =1 ⇔ tan x =1 ⇔ x = + kπ 4 cos3x + cos 2x − cos x −1 = 0 ⇔ 4t3 − 3t + 2t 2 −1− t −1 = 0
  17. t = ±1 cosx = ± 1 x= kπ    ⇔−1 ⇔ 2π ⇔ 2π t = cosx = cos x= ± + 2 kπ  2  3  3 1− sinx + 1 − cos x = 1 1− sinx − cos x + 2 (1 − sin x )(1 − cos x ) = 0 t 2 −1 ⇔ sin xcosx = 2 t 2 −1 1−t + 2 1 + −t = 0 ⇔−+=+t22142 t t 2 −−⇔− 24 t (1)0 t 2 =⇔= t 1 2 π  π   π   2 sinx +  = 1 sinx +  = sin    x= kπ 4  4   4  cos2 x sin x + +6 tan2 x() 1 − sin x = 2 1+ sin x t ∈[ − 1;1] 1−t2 t 2 t + +6() 1 −t = 2 ⇔ 6 t2 − t − 1 = 0 1+t 1 − t 2  π  x= + 2 kπ  1  6 t =  1  2 sin x =  5π ⇔ ⇔ 2 ⇔ x= + 2 kπ  −1   6 t = sinx = sinα   3 −1 x =arccos + 2kπ  3 1 sin6 x+ cos6 x = cos8 x 4 3 1 3 1− cos 4x  1 1− sin2 2x = cos8 x ⇔ 1 −   = cos8x 4 4 4 2 4 t ∈[ − 1;1]  2  π  πk π t= 4 x= + kπ x = + 3 1− t  1 2 2 4 16 4 1−  =()2t − 1 ⇔ ⇔  ⇔  4 2  4  − 2  3π  3πk π t = 4x= + kπ  x = +  2  4  16 4
  18. ToanDHSP.COM y 1 1  sin 2x + sin x − − = 2cot g2x 2sin x sin 2x  5x π   x π  3x  sin −  − cos −  = 2 cos  2 4   2 4  2 2  2cos x + 2 3 sin x cosx +1 = 3(sin x + 3 cosx) sin2x cos2x  + = tgx − cot gx cosx sinx 1  (2cos x −1)(sin x + sin2x) − cos 2x = 2  (2sin x +1)(2cos x −1) =1  sin3 x + cos3 x = 2 (1− sin x cos x) x  2sin x cos − cos x =1 2  π   π  3  sin4 x + cos4 x + cos x − .sin 3x −  − = 0  4   4  2 2sin x + cos x +1  = a a sin x − 2cos x + 3 1 a 3 a  x   tancx + os x − cos2 x = sin x1+ tan x tan   2  (2 − sin2 2x)sin 3x  tan4 x +=1 cos4 x
  19. Bài IV: Tích Phân ay eee y O ươ A • f( u( x)).' u( x) dx  a π 2 sin 2x I = ∫ 2 0 3+ cos x t=3 + cos2 x ⇒ dt=2cos x( − sin x) dx ⇒ dt= −2sin 2 xdx π 2 4 −dt 4 4 I=∫ = ln t⇒ I = ln 3 t 3 3 6 dx I = ∫ 2 2x+ 1 + 4x + 1 1 4x+ 1⇒ t2 = 4 x + 1⇒ tdt= dx 2 5 (t+1 − 1) dt 5dt 5 dt  1 5 3 1 ∫2 = ∫ − ∫ 2 =lnt + 1 + = ln − 3 ()t +1 3t +1 3 ()t +1  t +1 3 2 12 π 4 dx I = ∫ 2 0 cosx 1+ tan x
  20. ToanDHSP.COM dx 1t+an1 x ⇒ t 2 = + tan x ⇒ 2tdt = cos2 x π 4 2 2 2tdt 2 2 I = ∫ = 2 ∫ dt = 2t = 2 2 − 2 1 t 1 1 e 32−lnx I = ∫ dx. 1 x 1+ 2ln x dx 12+ln1 x ⇒ t 2 = + 2ln x ⇒ tdt = x e 2 2 2231− (t − ) 10 2 −11 Itd= ∫ t = ∫ (4 − t 2 ) dt = 11 t 3  aa  a ⇒  ⇒  du aa ∫ 2 (u ( x)) + a2  2 a2 + (u ( x)) ⇒ aa 2 a2 − (u (x)) ⇒ aa 3 dx ∫ 2 0 x + 9 a ⇒ dx =+3(tan2 t 1) dt π 4
  21. π 2 π 4 3() tant+ 1 dt 1 π I=∫ = t 4 = 9 tan2 t + 1 3 12 0 () 0 5 2 dx I = ∫ 2 1 9−()x − 1 ⇒ dx= 3cos tdt 5 2 π 6 π π π π 6 3costdt6 cos tdt6 cos tdt π I = ∫= ∫= ∫ =t 6 = 2 2 cost 6 0 9− 9sint 0 1− sin t 0 0 3 dx I = ∫ 2 2 1 x x + 3 3 tan t ⇒ dx=3( tan2 x + 1) dx π π 6 3 π 1 π 2 dt 3() tant + 1 1 32 −1 3 costdt I= dx =cos t = ∫ 2 2 ∫ 2 ∫ 2 3tant 3tan+ 3 3 π sint 1 3π sin t 6cos2t cos 2 t 6 π π 1 3 d()sin t 13 6− 2 3 I = −∫ 2 = − = 3π sint 3sin t π 9 6 6
  22. ToanDHSP.COM Ư b b b ∫udv =−uv ∫ vdu a a a y aayaya  ∫ P( x)ln xdx a ln x y eax+b     ∫ P( x).sin(ax + b) dx a   cos(ax + b) yyaya π 2 I =+∫ (x 1)sin2xdx. 0 u = x +1⇒ du = dx π π  −( x +1) 1 2 π  −1 ⇒ I =+=+ cos 2x 2 cos 2xdx 1 ⇒ ∫ dv = sin2xdx v = cos 2x 224 0  2 0 2 I = ∫ (x − 2)lnx dx. 1  1 du = dx u = ln x  x  x2  2 2  x  5 ⇒ ⇒ I =−− 2x ln x − 2 dx = −ln 4 +   2   ∫  dv = ( x − 2) dx x  2  1  2  4  v = − 2x 1  2 π 2 4 ∫ sin xdx 0 x ⇒ t 2 = x ⇒ 2tdt = dx π 2 4 π 2
  23. π 2 B= 2∫ t sin tdt 0 π 2 I= ∫ tsin tdt 0 u= t du= dt  ⇒  dv= sin tdt v= −cos t π π π 2 π π I= − tcos t2 + cos tdt = − cos + 0cos 0 + sint 2 = 1 ∫ 2 2 0 0 0 π 2 ∫ ex cos xdx 0 u= ex du= ex dx  ⇒  dv= −sin xdx v= − cos x π π π π 2 π π 2 2 A= − excos x2 + e x cos xdx = − e2 cos + e0 cos 0 + ex cos xdx = 1 + e x cos xdx ∫ 2 ∫ ∫ 0 0 0 0 π 2 K= ∫ ex cos xdx 0 u= ex  du= ex dx  ⇒  dv= cos xdx v= sin x π π 2 π K= exsin x2 −∫ e x sin xdx = e2 − A 0 0 π π π 1+ e 2 ay A=1 + e2 − A⇒ 2 A= 1 + e2 ⇒ A = 2 π ∫ xsin x cos2 xdx 0 u= x du= dx ⇒ 2  2 dv= sin x cos xdx v= ∫sin x cos xdx v= ∫sin x cos2 xdx t= cos x⇒ dt= − sin xdx
  24. ToanDHSP.COM −t3 cos3 x − t 2dt = + C = − + C ∫ 3 3 cos3 x ⇒ v =− 3 cos3 x π 1 π π 1 y A = −x + ∫ cos3 xdx = + K 3 0 3 0 3 3 π π K = ∫ cos3 xdx = ∫ (1− sin2 x)cos xdx 0 0 ⇒ dt = cos xdx π 0 K = ∫ (1− t 2 ) dt = 0 0 π 1 π ay A = + K = 3 3 3 π 2 x + sin x D = ∫ dx π 1c+os x 3  π ux=x + sin  2  du = (1c+os x) dx x + sin x  1  D =  ⇒  ∫ x dv = dx x π 2  2 x v = tan 2cos 2cos  2 3 2  2 π π x 2 2 x  ππ   33 y D = ( x + sintxan) 1cos − ∫ ( + x) tan dx =  +1 −  +  − K 2 π π 2  2   3 2  3 3 3 π π π 2 x 2 x x 2 K = ∫ (1+ cos x) tan dx = ∫ 2cos2 tan dx = ∫ sin xdx π 2 π 2 2 π 3 3 3 π 2 1 = −cos x = π 2 3 (9 + 2 3)π aya 18 aaaaa ưư y
  25. y π 3  I= ∫sin2 x . tgxdx 0 7 x + 2  I= dx ∫ 3 0 x +1 e  I= ∫ x2 ln xdx 0 π 4  I=∫ ( tgx + esin x cos x ) dx 0 π  I= ∫ cos x sin xdx 0 π 3  I=∫ tan2 x + cot 2 x − 2 dx π 6 π 2  I =∫ 2() 1 + cos 2x dx −π 2 π 3 sin 4x sin 3 x  I = ∫ dx π tanx+ cot 2 x 6 10 dx  I = ∫ 5 x− 2 x − 1 e 3− 2ln x  I = ∫ dx. 1 x 1+ 2ln x π xsin x  I = ∫ 2 0 1+ sin x π 6 sinx+ sin3 x  I = ∫ 0 cos 2x  aa (P) : y= x2 − x + 3 d : y= 2x + 1. x2 27  ()()C1 y= x2 ; C 2 y = ;() C 3 y = 27 x
  26. ToanDHSP.COM a a ya yay aaaaa yy aay A y = f ( x) f ( x) ≥ 0;∀x ∈ I  f ( x) ≤ 0;∀x ∈ I  1 y = f ( x) = x3 − mx2 + (m2 + m − 2) x 3 a (4;+∞) e (−∞;4) (2;+∞) D = R y ' = x22− 2mx + m + m − 2 ⇒ ∆ ' = −m + 2 a  ∆ ' ≤ 0 ⇔ −m + 2 ≤ 0 ⇔ m ≥ 2 y '0(0) ≤ m2 + m − 20≤   ⇔  ⇔ m ≤1 2 y '(2) ≤ 0 m − 3m + 2 ≤ 0  ∆ ' ≤ 0 ⇔ −m + 2 ≤ 0 ⇔ m ≥ 2
  27.  ∆' > 0 m 0   −m +2 > 0  m 2 y '() 4≥ 0  2 ≥ ⇔ ⇔ m+9 m + 14 ≥ 0 ⇔    −2 ≤m ≤ 1  y '()− 2 ≥ 0 2   m−3 m + 2 ≥ 0  S  −4 < < 2 −4 <m < 2  2 −1 m2 y= x2 + mx 2 +() m − m2 x + 3 3 a (6;+∞) e (−∞;0) (6;+∞) y'= − x2 + 2 mx + m − m2 ∆' = m a ⇔ ∆' ≤ 0 ⇔m ≤ 0 y '( 0) ≥ 0 −m2 + m ≥ 0 ⇔ ⇔  ⇔m =1 2 y '() 2≥ 0 −m +5 m + 4 ≥ 0 (6;+∞) ∆' ≤ 0⇒ m ≤ 0 a (6;+∞)
  28. ToanDHSP.COM  ∆>'0 m > 0   ym'(m6) ≤ 0 ⇔ − 2 +13 − 36 ≤ 0   S m < 6  < 6   2 m ≤ 0 y ⇔  ⇔ m ≤ 4 m∈[0;4] 2 ∆ ' ⇔ x − x = 2 ⇔ = 2 ⇔ 2m = 2 ⇔ m =1 1 2 a e (−∞;0) (6;+∞) ∆ ' ≤ 0 ⇔ m ≤ 0 ∆ ' ≥ 0  y '(0) ≤ 0  y '(6) ≤ 0 ⇔ 1≤ m ≤ 4   S 0 << 6  2 m ≤ 0 y  1≤ m ≤ 4 A 1 y = x3 − mx2 + (2m2 −1) x + m3 − m 3 a y
  29. e a 3 3 a ( x1+ x 2 ) y'= x2 − 2 mx + 2 m 2 − 1 ∆' = −m2 + 1 ∆' > 0 −∞ +∞ a  y '( 0) = 0 2  2m − 1 = 0 2 ⇔S ⇔  ⇔m = 0 0 2  2   m 0 −m2 +1 > 0    2 m 0 ⇔  2 m − 2 m > 0 ⇔  ⇒ −1 1 S m 0 m2 0 ⇔y'() − 1 > 0 ⇔ 2 m + 2 m > 0 ⇔ ⇔ 0 −1  > −1  m > −1  2  ∆' > 0  m <1   y '()− 2 ≥ 0 2  2m+ 4 m + 3 ≥ 0() ∀ m ⇔ y '() 3≥ 0 ⇔  ⇔ −1 <m < 1  2m2 − 6 m + 8 ≥ 0() ∀ m  S  −2 ≤ ≤ 3 −2 ≤m ≤ 3  2 e
  30. ToanDHSP.COM −1 '0 −1 0 m ≥ 0 0 a y '0(0) 0 ⇒ m 0 ⇔  3 P = ( x12+1 x ) − 3x x2 ( x1 + x2 ) →min 2 xxm = 2 2 −1 −+m> 10  1 2 ya ⇔  x x m 3 2  1 += 2 2 Pm=m−(−2m→) 3(2 1).2 min −1 ' 0
  31. S −∞ +∞ 2 y '(− 2) ≥ 0 a  y y '() 3≥ 0 yaaaa af (α ) ya aa f (α ) y aeaa S −b aayay 2 2a S −b y a 2 2a ay yy ayaye y aaayyy ya y ∞ yye aay eye yya a ax2 + bx + c  y = a''' x2 + b x + c a b a c b c x2 +2 x + a'''''' b a c b c y ' = 2 yyay (a'' x2 + b x + c) a ≥ a y  ya ax+ b aaya y = a'' x+ b y= x3 −3 mx 2 + 3( m − 1) x + 4 a aa aa 2 5 aa ∆:y = 2
  32. ToanDHSP.COM y ' = 0 aa  y = f (x) y y ' = x2 − 2x − m +1 = 0 y = x3 − 3mx2 + 3(m −1) x + 4 ⇒ y =−−x+2 ++2+1x m (cx d ) ax b = ax + b () 0 ⇔ y = ( x2 − 2x − m +1)(x −1) − 2mx − m + 5 x2 + 2x − m +1 = 0(1) ⇔  y = −2mx − m + 5(2) a ⇒ ∆≥' 0 ⇒ m > 0 a aa ⇒ aa y = −2mx − m + 5 ⇒ ∈ ⇔ m = 2 y aa 2 5 2 ∆ ' (1) ⇒ x −=x= 2 m 2 1 a 2 2 (2) ⇒ y2 −=y−1 2m( x2 − x1 ) = −4m m ⇒ AB = ( x2 − x1 ) + ( y2 − y1 ) = 2 5 m =1 2  ⇒16mm+=4⇔ 20 5 ⇒ m =1 m = −  4 aa ∆ : y = 2 ⇔ d ( A;∆) = d (B;∆) ∆ : y = 2  y1 − 2 = y2 − 2  y1 = y2 ⇔ y1 − 2 = y2 − 2 ⇔  ⇔   y1 − 2 = −( y2 − 2)  y1 + y2 = 4 ⇔ (−2mx1 − m + 5) + (−2mx2 − m + 5) = 4 ⇔ −2m( x1 + x2 ) − 2m +10 = 4 ⇔ −2m.2 − 2m +10 = 4 ⇔ m =1 y = x3 + 3x2 − 3(m −1) x a aa ∆OAB aa aay ay e aaa 2 2 aa ( x −1) + ( y −1) = 4 aaaaa aaaaa
  33. y '= 0 aa  y= f() x  y ' 2  =x +2 x − m + 1 = 0 ⇔  3  3 2 2 yxx=+−3 3() mxxxm −=+−+ 1( 2 1)() x +− 1 2 mxm +− 1 x2 +2 x − m + 1 = 0( 1) ⇔  y= −2 mx + m − 1() ∆ a ⇒ ∆' ≥ 0 ⇒ m > 0 a aa ∆OAB   ⇔OA ⊥ OB    OA= ( xAA; y ) ⇔ OAOB.  OB= () xBB; y ⇔xxyy1 2 + 1 2 =⇔0 xx1 2 +−( 2 mxm1 +−− 1)( 2 mxm2 +−= 1) 0 2 2 2 ⇔x1 x 2 +4 m x 1 x 2 +( − 2 m + 2 m)()() x 1 + x 2 + m −1 = 0 2 ⇔ ()()−++m1 4 m2 −++− m 1( 2 m 2 + 2 m) . −+−= 2() m 1 0 ⇔ −4m3 + 9 m 2 − 7 m + 2 = 0 ⇔( −4m2 + 5 m − 2)( m − 1) = 0 ⇔m =1 a  VN vì ∆=− 7 y CD aa ⇔y1. y 2 ⇔  4 m ≠ 1 aay ⇔x1 x 2 > 0 x1 x2 ⇔ −m +1 > 0 ⇔ m < 1 ay  x+ x   y a  1 2 ;− 2mx + m − 1  2  ⇒ M(−1;3 m − 1) Ycbt⇔5 = 3 m − 1 ⇔ m = 2 ay e aaa ∆:y =− 2 mx +−⇔∆ m 1 : 2 mx +−+= y m 1 0
  34. ToanDHSP.COM 2.m001+ − m + 2 2 ⇔ d (O;∆) =1 ⇔ =1 ⇔ (−m +1) = (2m) +12 ⇔ 3m2 + 2m = 0 (2m)2 +12 m = 0  ⇔ −2 ayay m =  3 2 2 aa ( x −1) + ( y −1) = 4 ⇔ d (I;∆) = R ∆ : 2mx + y + m −1 = 0 2.m111+ − m + 2 ⇒ = 2 ⇔ (m + 2) =16m22+ 4 ⇔ −15m + 4m = 0 (2m)2 +1 m = 0  4 ⇔ 4 a m = m = 15  15 aaaaa −2mxm+ −1 = 0  m −1  aa ∆ ⇒  ⇒ M  ;0 y = 0  2m  ym=m−+2 .0 −1 aa ∆ y ⇒  ⇒ Nm(0; −1) x = 0  m =1  m −1  1  1  ⇔ xM = yN ⇔ = m −1 ⇔  −1. m −1 = 0 ⇔ m = 2m  2m   2  −1 m =  2 −1 1 y ∆ aa aya m = 2 2 aaaaa 1 1 1 ⇔=S⇔= OM.ON x y ∆OMN 2 8 2 M N  m = 2  2 m  2 m − 2m +1 = ⇔ 1 1 m −1 1 (m −1)  2 m = ⇔=− . m 1 ⇔ = ⇔   2 4 2m 4 2m  −m m2 − 2m +1 = (VN )  2 yaa
  35. ƯAOAA C1 :;: y= f( x) C2 y= g( x) aaaa f( x) = g( x)  f( x) = g( x)   f''()() x= g x ưuaa aya 2mx− 3 m − 2 ()C: y = ()m ≠ −2 (d) : y= x − 1 m x −1 a a [−2;3] e aa 2mx− 3 m − 2 =x −1 ⇔ g()() x : x2 − 2 m + 1 x + 3 m + 3 = 0 x −1 S −∞ x x +∞ 1 2 2 g( x)  m > 2 ∆' > 0 ⇔  aa  m 0  −6 −6  1+ 2()m + 1 + 3 m + 3 > 0 m >  − 1    2 m > −2 m > 2 g (2) > 0  4− 4()m + 1 + 3 m + 3 > 0 −m +3 < 0 m < 3 S ⇔  ⇔  ⇔   < 2 m +1 < 2 m<1  m <1  2
  36. ToanDHSP.COM m 0  3mm+ 3 > 0 ⇔ > −1 ⇔  S ⇔  0 ≤ mm+1≥ 0 ⇔ ≥ −1  2 aya e g (0) 0
  37. y ưuy ∆ aa ∆ x a aa x1 0 m2 +8 m 0 m. g () 1> 0 −2m > 0  m 0 k > 0 k > 0  ⇔  ⇔  a g ()2≠ 0 4+ 4 −k + 1 ≠ 0  k ≠ 9 B( x1;;; y 1 ) C( x2 y 2 ) a 2 2 2 x+2 x − k + 1 = 0( 1) x  y= kx −2 k + 4() 2 2∆ ' ⇔x − x = = 2 k 2 1 a Hình 3 ⇔y2 − y 1 = k( x2 − x 1 ) = 2 k k  2 2 BC=()() x2 − x 1 + y 2 − y 1 = 2 2 ⇔4k + 4 k 3 = 2 2 ⇔ 4k3 + 4 k − 8 = 0 ⇔ k = 1
  38. ToanDHSP.COM y ∆ : y =1( x − 2) + 4 y = f ( x) = x3 −3x2 + 2 y a ay ayya a ay A(a;−2)∈ d : y = −2 ∆ a A(a;−2) yk=xa ( − ) − 2 (∆) ∆ a x3 − 3x2 + 2 = k ( x − a) − 2 (1)  2 3x − 6x = k (2) aya x3 − 3x2 + 2 = (3x2 − 6x)( x − a) − 2 (3) ⇔2x3 −3(a +1) x2 + 6ax − 4 = 0 3 ⇔ ( x − 2) 2x − (3a −1) x + 2 = 0 x = 2 ⇔  2 gx(xa)x=−2− (3 1) + 2 = 0 (4) ay   2  5 ∆g > 0 (3a11−60) − > a ⇔  ⇔  ⇔  3 (*) g (20) ≠ 2.22 − 3a −1 .2 + 2 ≠ 0  ( ) a ≠ 2 ayya x0 = 2; x1; x2 aay 2 2 k0 = f '(2) = 0; k1 = f '( x1 ) = 3x1 − 6x1; k2 = f '( x2 ) = 3x2 − 6x2  k0 = 0 2 2  2 2  ⇔ (3x1 − 6x1 )(3x2 − 6x2 ) = −1 ⇔ 9 x1 x2 − 2x1x2 ( x1 + x2 ) + 4x1x2  = −1 ( ) ea 3a −1 x + x = x x =1 1 2 x 1 2  3a −1  55 ⇔ 9 1− 2  + 4 = −1 ⇔=a a   2   27  55  y A ;−2  27 
  39. OƯOƯ ươ (Cm ) y (Cm ) (Cm ) ya n a y= f( x; m) y= ±()()( axbm + + gx nnguyn: ê≥ 2) (C) : y= g( x) n ±()()()ax + bm + g x = g x ∀m  n−1 ±na()()() ax + bm + g'' x = g x (Cm ) y ∆ a  c ax+ b + = k()() x − x0 + y 0 1  x+ d   c a−2 = k()() x ≠ a 2  ()x+ d a a c ax+ ad − = k()() x + d 3 x+ d 2c 2c b− ad + = k() − x − d + y ⇔ =k() − x − d + y + ad + b ()4 x+ d 0 0 x+ d 0 0 ayeya ∀m ax+ b cx+ d 3 2 2 (Cm ) : y= x + 2 x +( 2 m + 1) x + m + 2 3 2 2 2 3 2 a (Cm ) : y= x + 2 x +( 2 m + 1) x + m + 2 ⇔()x + m + x + x + x + 2 (C) : y= x3 + x 2 + x + 2 (Cm ) a 2 ()x+ m + x3 + x 2 + x +2 = x 3 + x 2 + x + 2  1 2 2 () 2()x+ m + 3 x + 2 x + 1 = 3 x + 2 x + 1
  40. ToanDHSP.COM 2 ( x + m) = 0 a (1) ⇔  2( x + m) = 0 3 2 y ∀m (C) : y = x + x + x + 2 (m − 2) x − (m2 − 2m + 4) (C ) : y = a m x − m 2 (m − 2) x − (m − 2m + 4) 4 (C ) : y = ⇔ ym= ( − 2) − m x − m x − m (∆) : y = ax + b ⇔ a ∀m  4 (ma−x−2b=1)+ ( )  xm−  (I )  4 2 = a (2) ( xm− ) a 4 ⇒ =−a ( x m) (3) x − m y 8 −8 ⇔ (m − 2) − = b + am ⇔ = (a −1) m + b + 2 (4) x − m x − m ay 2 ⇔ (a −1) m + (b + 2) =16a 2 2 ⇔ (a −1) m2 + 2(a −1)(b + 2)m + (b − 2) −16a = 0 (*) ∀m ⇔ (*) ∀m (a −1)02 =  a =1 ⇔ 21(2a0− )(b + ) = ⇔  b = 2 ∨ b = −6  2 (b +−2=) 16a 0 yyy
  41. y 1 1 y= x3 −() m +1 x2 + 2() m 2 + m x − 3 3 a 3 e 3 2 2 3 (Cm ) : y= x + 3 mx + 3( − m + m + 1) x + m + 1 a x1 x 2 −14 a + = x2 x 1 5 3 (Cm ) : y= x − 3 x + 2 a y ya y ay ay aya y ay ay e y 4 2 (Cm ) : y= x − 2 mx + 2 m − 1 y x3+ mx 2 −1 = 0 3 2 2 3 (Cm ) : y= x − 3 mx + 3( m − 1) x − m 3 (Cm ) : y= x + k( x +1) + 1 (∆) :y = x + 1 3 2 3 (Cm ) : y= x − 3 mx + 4 m (d) : y= x a 2x + 1 ()C: y = ya m x + 2 (3m+ 1) x − m2 + m ()C : y = ()1 a m x+ m a a∀m a 3 2 ( Cm ) : y= ( m+−+−+++ 3) x 3( m 1) x( 6 m 1) x m 1( 1) a
  42. ToanDHSP.COM yy yyay yy 5 − x32− 3 x + 7 lim x→1 x2 −1 5 − x323− 3 x + 7  5 − x − 2 3 x2 + 7 − 2  a lim = lim −  (1) x→1 2 x→1  2 2  x −1  x −1 x −1  2 5 − x3 − 2 1− x3 −( x + x +1) −3 lim 2 = lim lim = (2) x→1 x −1 x→1 ( x2 −1)( 5 − x3 + 2) x→1 ( x +1)5(2 − x3 + ) 8 3 x2 + 7 − 2 x2 −1 11 lim 2 = lim lim = (3) x→1 x −1 x→1  2  x→1 22 2 3 12 ( x2 −1) 3 ( x2 + 7) + 2 3 x2 + 7 + 4 3 ( xx ++7+2)+74   −3 1 11 ay A = − = 8 12 24 ư aa a a ayaya a  57− x32− c 3 x + − c  ∀c ∈ R f x =  −  ( )  22   x −11 x −  3 aa f1 ( x) = 5 − x − c 3 2 f2 ( x) = x + 7 − c yaay   f1 (1)0=  c = 2  f2 (1)0=   ⇔ c = 6 ⇔ c = 2 a   f1 (−=1)0   c = 2  f2 (−1=)0 yyy aaa f ( x) 0 F ( x) = g ( x) 0
  43. f( x) + c f( x) − c f() x =1 + 2 g() x g() x αi (i =1;2; ) a  f1 (αi ) + c = 0 a  ()i =1;2;  f1 ()αi − c = 0 f( x) + c f( x) − c lim 1 lim 2 e x→αi g() x x→αi g() x ay 33x 2 − 1 + 2 x2 + 1 lim x→0 1− cos x 1+ 2x −3 1 + 3 x lim x→0 x2 ơơo yyy ayaa yyay a ay • a log x b • a loga x = logb a m logxn = log x an n a  • t= loga f( x) f( x) t= a  • 2 2 2x2− 3 x + 1 81.42x− 3 x + 1− 78.6 2 x− 3 x + 1 + 16.9 ≤ 0() 1 2 2 2 2 6 2x− 3 x + 1 9 2x− 3 x + 1 3 2x− 3 x + 1 3 2.( 2x− 3 x + 1) ()1⇔ 81 − 78 +16   ≤ 0 ⇔81 − 78 +16   ≤ 0 4  4  2  2  2 3 2x− 3 x + 1 t =   2  3 27  16t2 − 78 t + 81 ≤ 0 ⇔t ∈ ; 2 8  2 3 3 2x− 3 x + 1 27 ⇔ ≤  ≤ ⇔1 ≤ 2x2 − 3 x + 1 ≤ 3 2 2  8
  44. ToanDHSP.COM  3 x ≥  2 2 2 x ≥ 2 2xxx−x3 +1 ≥1 2 − 3 ≥ 0 x ≤ 0   ⇔  ⇔  ⇔ 1 2xx2 x−x3 +1 ≤ 3 2 2 − 3 − 2 ≤ 0  1 x ≤ x ≤  2  2  x ≥ 2 ex+ x−1 − e1+ x−1 ≤ x −1 u = x + x −1  ⇔ u − v = x −1 v =+1− x 1 eu − ev = u − v ⇔ f (u) ≤ f (v) f ( x) = ex − x; x ≥ 1 ⇒ f '( x) =+e>x 1 0 ⇒ f ( x) u ≤ v ⇔ x + x −1 ≤ 1+ x −1 ⇔ x ≤ −1 log2 (1+ x ) = log3 x t log3 x = t ⇔ x = 3 t t t log2 (1+ x ) = t ⇔ 1+ x = 2 ⇔ 1+ ( 3) = 2 t t 2  1 t  3   1 t  3   1 2  3    +   = 1 ⇔   +   =   +    2   2   2   2   2   2  t  1 t  3  ⇔ f (t) = f (2) ⇔ t = 2 fx( ) =   +    2   2  ⇔ t = 2 ⇔ x = 9  x+1 log x log2 (4 −≥8) 1 (1) x−1 2(x−1) 3 5 4 −8 > 0 ⇔ 2 > 2 ⇔ 2( x −1) > 3 ⇔ x > 2  x+1 x−1− x 1 x (1) ⇔−lo≥gxxlog2 (4 8) log x ⇔−lo≥g2 (4 8) x ⇔ log2 (4 −8) ≥ log2 2 4x 20x ≤ (loai) ⇔ 4x−1 − 8 ≥ 2x ⇔ − 2x −8 ≥ 0 ⇔  ⇔ x ≥ 3 4 28x ≥  x −1 + 2 − y =1 (1)  2 3 3log9 (9x ) − log3 y = 3 (2)
  45. x ≥1  0 1  x + 2 2 2x2 − x − 1 ⇔ =2v à x > 1 x + 2 5 ⇔2x2 − 3 x − 5 = 0 v à x > 1 ⇔ x = 2 2 2 2 log2 ( x+ 6 x − 7) logx+ x − 15logx − x − 452 = ≥ 2 3 ( ) 5 ( ) 1  2 1+ log4 x − x +  log( x− 2) + log x ≥ log( x + 2) 4  0.2 3 5 2 ( x+3) log2( x + 2) + 4( x + 2) log 3 ( x + 2) = 16 log2 x+ log3 x ≥ log2 x log 3 x xlog2 3 + x2 = xlog2 5 1 log2 () 3x − 1 + =2 + log2 ()x + 1 logx+3 2 aay ex− e y =ln( 1 + x) − ln( 1 + y)  x− y = a  2 2 log2( x+ y ) =1 + log 2 ()xy  x, y∈ R 2 2 () 3x− xy + y = 81
  46. ToanDHSP.COM a a x x sin2 x cos2 x ( 5 +1) + 2m( 5 −1) = 2x 9 + 9 = m log 2 (3− x) >1 73x + 9.52x = 52x + 9.73x 3x−x x−3 x−3 x x 7 5 16 + ( x − 6) 4 + 8 − 2x = 0 (5) = (7) x x−10 5 10 3 + 3 − 84 = 0 log x2 +1 < log ax + a 1 ( ) ( ) 22( ) ( ) ( ) x−3 x−3 16 + ( x − 6) 4 + 8 − 2x = 0 a a aa x2 −x x − 3 = 9 − 6x + x2 2(5x + 4) − 5x − 3 ≤ 5x + 3 3.25x−2 + (3x −10).5x−2 + 3− x = 0 log2 ( x + y) + logm ( x − y) =1  x22− y = m (m + 3)16x + (2m −1)4x + m +1 = 0 9x − m.3x + 2m +1 = 0   3x −1  x +1    log3 log4   ≤ log1 log 1  x 15  x +1  3x −1 log2 log0.5  2 −  ≤ 2   3  4    16 
  47. Aea U 1 y=()() x2 − m x2 +1 a 4 ay 7 cos23 x+ sin2 x = 2sin x 2 x x+( x −4) 4 − x = 4( x − 2) x2 + log( cos x) lim 2 x→0 2xsin x −x 2 + 1 aaaa ya ∆ 2 2 2 y∈[0;1] A=()()() x − y + y − z + z − x ươu ay ∆ C∈ d: x − 3 y + 1 = 0 a a x y−1 z − 1 y ∆ a d : = = 1 −1 1 1 x+1 y − 2 z − 4 d = = = 2 2 1− 1 n n−1 n an ()()() x−1 + an−1 x − 1 + +a 1 x − 1 + a 0 = x , ∀ x ∈ R a2+ a 3 + a 1 = 231 ươa x2 y 2 ay ME∈() : + = 1y 6 3 yaya 1 1 1 1 + + = OA2 OB 2 OC 2 OM 2 2n n nπ a ()1+ i y CCC0− 2 + 4 − +() − 1C 2n = 2 n cos 2n 2 n 2 n 2n 2 (n∈ N, n > 0)
  48. ToanDHSP.COM U 2 y = −x4 + x2 9 ya (1− 3 cos x)sin x + ( 3 − cos x)cos x =1  x + 2y − x − 2y = 2  2 2  3 x + 3 + x − 4y = 5 e dx ∫ 2 1 x + x 1− ln x a ∆ (a < b < a 2 ) x, y ∈[1;2].  1 1   1 1  2 2 Ax=yx(y + ) 2 + 2  + 4( − ) −   x y   x y  ươu 2 2 x y ay M ∈(E) : + =1 6 3 ya ∆ ya x − 2 y z −1 x −1 y +1 z d : = = d : = = a 1 −2 1 1 2 1 −2 1 e ươa ayyeaa yy 2 3 + i a zn y
  49. U mx + 2 y = x+ m yay 1 cos2x+ cos x + 3sin x = 2 logx+ x2 − 12.logx − x2 − 12 = 2 2 ( ) 3 ( ) π 2 sin 3xdx ∫ 4 0 ()1+ cos x yaa  y2 1 x2 − xy + = a  2 4  2 x+ x − y = m ươu aaaya 4 2 y x− 3 y z ya d : = = 4− 5 3 a a b 5i a, b∈ R + = 3 z = z+1 z − 5 1 1+ 2i ươa aa ∆ y A∉ Ox; B ∈ Oy C∈ d: x + y − 1 = 0 aaay y a
  50. ToanDHSP.COM U x3 y = − + x2 3 aa a ya 4tan x + 2tan x+sin 2x = 21+2sin 2x 2 + 2x − 3 x ≥ x 2 − 2x − 5 x π 4 sin4 x ∫ 44 dx 0 sin x + cos x yaa a2 + b2 + c2 aa A = a3 + b3 + c3 − 3 ươu yy aaa AI + 2BI = 0 M ∈(P) a yaa ươa ayaya ya a ∆ x2 − (m +1) x + 2m −1 y = x − m
  51. U x − 3 y = x +1 a 3 cos2 x− cos x cos3 x + cos32 x = 4  1 1  x2 + + y2 + = 3  x2 y2   1 1 + =1  x+ y xy 4 3 xln x dx ∫ 2 3 1+ x 4 ⊥ ∆ ∆ a 2 1 1 a b25 ab a + =1 A = + + a b a−1 b − 1 4()a2+ b 2 ươu yy aaa ∆ MP∈() a a b a, b∈ R Z= i − i2 + i 3 − i 4 + + i 2009 a + =1 1+z 1 − z ươa x= t  x−1 y z y d1 : y= 1 + 2 t d2 : = =  1− 1 1 2 + t a A∈ d1 6 a x  log3 log3 2 2 +y = 6 logy+ log x = 1  x y  x3
  52. ToanDHSP.COM U x4 y = − mx2 + m +1 a 4 aa  π   π   π  sin  2x +  + cos 2x +  = tan  x +   6   3   4   3 3  1 1   1 1  ( x + y ) 3 + 3  + ( x + y) +  = 8   x y   x y    x y log log =1  2 2 3 3 3 xdx I = ∫ 1+x2 0 e aa 1 1 1 b + c c + a a + b a + + =1 A = + + a b c b3 + c3 c3 + a3 a3 + b3 ươu ay ∆ A∈ d1 = 2x − y +1 = 0 B,C ∈ d2 : x + y − 2 = 0 a ya (Q) : x − y − 2z = 0 1 a C3 − C 2 = A3 n+1 n+1 7 n ươa 2 2 2 2 a y a (Cm ) : x + y − 2mx − my + m − 2 = 0 (C) : x + y − 3x +1 = 0 yaa y ( P) : x + 2y + z −1 = 0 x − 2 y +1 z ∆ : = = a 2 1 −1 z4 + z2 +1 = 0 z ∈C
  53. U y= x3 −3 ax 2 + b (a, b > 0) a aa ∆  x  2 tan2x 1+ tan x .tan  =  2 sin 3x  1 1 1  + =  x+ y xy 2  5 2 1  − =  x2+ y 2 x 2 y 2 2 ex − x +1 lim x→0 ln() 1+ sin x aayaa aaaa 1 +ln ( x + x − m +2 x − m ) ≤ 1 2 mx− x2 ươu ay ABC(7;1,) (− 3; − 4,) ( 1;4) ∆ x−1 y + 1 z − 2 yaa d : = = 1 2− 1 x+1 y − 2 z ∆: = = 2 1 1 6 a x3 aaa( x2 + x −1) ươa ay (C) : x2+ y 2 − 6 x + 5 = 0 aaa yaaay x−1 y + 1 z y M (2;1;0) = = −2 1 − 1 aa 5 a x3 aaa( x2 + x −1)
  54. ToanDHSP.COM U mx +1 y = x +1 ya sin2 x + sin 2x.sin 4x = cos2 2x 22+3x + 22−3x − 7(2x + 2− x ) ≤15 1 1 aya (C) : y = 1+ + 1− , x x aya aa d ;d a   1 2 y M ∈d1, N ∈d2 a AM ,CN a  11 2 xy + = x +  x 1+ y ln y  1 1 xy + = y2 +  y 1+ x ln x ươu aya (P) : y2 = x a ∆ a ( yA < 0) x −1 y −1 z − 2 yaa d : = = 2 2 1 aaaa a ươa ayeaa MF1.MF2 yaa (Q) : x + z −1 = 0 và (R) : x + 2y − z +1 = 0 Z = (1+ 2i + 3i2 + + 2009i2008 )(1− 2i + 3i2 − 4i3 + + 2009i2008 )
  55. U y=( x − m)( x2 − x +1) yaaya 1− sinx + cos x tan x = 1+ sinx + cos x 2 log2 log2x+ log 2 x 0.25 ()7+ 5 2x =() 3 − 2 2 (C) : y= x2 − 2 x − 3 d: y= x + 1 aaya ⊥ a 1 1  +2 = 3 a  x y y  2 2x+ y = m ươu ay ∆ ay hA =2 x + y + 4 = 0 mA = y −2 = 0 y mB : 3 x+ 11 y + 21 = 0 x= t x−2 y − 1 z − 2  y d1 : = = ,d2 :y = 2 t a 1 2 1  z=1 + t aa x, y∈ R a 1 1 1 − = x+()2 − y i 2 + y + xi ()1+ i 2 ươa x2 y 2 ay ()H :− = 1 ()a , b > 0 a F ; F a; F a2 b 2 1 2 2 a ∆F1 MN aa ∆F1 MN = 4 3 yya ∆ 3x 3 y = + 4 x ∆
  56. ToanDHSP.COM U −x4 y = + (m +1) x2 − m, (1) a 2 ay a  π  4sin 2xsin  x +  =1+ 3 sin 2x − cos 2x  3  x2 − xy + 4y = 8  xy + y2 + 3x = 12 x y = a e x+ln x a aaa   log2 ( 2x − 4 + m) =1+ log3 m − ( x −1)( x − 3) y y ươu x + 2 y z −1 y d : = = y 2 1 1 aaaa aa a 6 3 x3 1 log323 log x −=lo+g  log2 x  x   3  2 ươa xt=  yaa d1 : y = 1+ 2t y  2 + t a ∆ a a 2π 2 x2 + (2 + cosϕ) x + 3 sinϕ φ ∈(0;2π ) y = a x −1 a
  57. U 2x y = , () 1 x −1 a yaa 16sin2 x+ 4cos 4 x = 3 cos x + sin x 3 x−5( x − x − 5) = 2 2 2 aya ()()()C: x− 3 + y − 1 = 1aya y a a 2 a a a2+ b 2 + c 2 =1. a3+ b 3 + c 3 −3 abc ≤ 1 ươu ayay a∆ y∈ayπ 0 1 2 x−2 CCCC2 +3 + 4 + +x = 120 , x ∈ N ươa yy   a∈a AM+ BM aeay  x x2− y 2 x− y  y xy xy 4 +2 = 5.4  log3 x+ log5 y = log5 x .log 3 y
  58. ToanDHSP.COM U −x3 16 y = + mx2 − 2(m − 2) x + (1) 3 3 sin 3x + sin x = 3 (cos x −1) 4 log + log 0, 25 ≥ log x2 2 x2 2x 0.5 1 4 x I = ∫ dx 0 12− x ayayy aaaa 1 1 a2 + b2 ab a + =1 P = + a + b ab ab a + b ươu ay ∆ ∈y 3x − y −10 = 0 a yA > 0 > yB yya 2 2 4z 4 +1 = 0 ươa ay (P) : y2 = 2x a F aya F 1 1 + MF NF yaa ⊥ y n+1 n a x6 aaa P = ( x +1) ( x2 + x +1) a x10
  59. aa ay yaaay ayay yyeeyy 2x3 − 3( a + 1) x2 + 6 ax − 4 = 0( 1) y 2x3 − 3( A + 1) x2 + 6 Ax − 4 = 0 aya aa yayayy ya y ayayy ya ya a y= x3 −3( m + 1) x2 + 2( m 2 + 4 m + 1) x − 4 m( m + 1) x3 −3( A + 1) x2 + 2( A 2 + 4 A + 1) x − 4 A( A + 1) = 0 y yaaa a yaayya a ya yay a ay a sin 2x+ cos 2 x − cos x + 3sin x = 2( 1) yaa ay y π π π a ±; ± ; ± 3 6 2 π π và yaaa 2 6
  60. ToanDHSP.COM 1 y a 2 (sin x −1) ay (2sin x −1) ae (sin x −1) (1) ⇔ 3sin x − 3+1− cos x + sin 2x + cos 2x = 0 ⇔ 3(sin x −1) +1+ (1− 2sin2 x) + sin 2x − cos x = 0 ⇔ 3(sin x −1) + 2(1− sin2 x) + sin 2x − cos x = 0 ⇔ (sin x −1)(1− 2sin x) + 2sin x cos x − cos x = 0 (sin x −1)(1− 2sin x) + cos x(2sin x −1) = 0 ⇔ (sin x −1)(1− 2sin x + cos x) = 0 yaa y OE x = 2 + 2t x =1   y d1 : y = −1+ t d2 : y =1+ t   z =1 z = 3− t a a a eayae eyaeA aeayae e aaae  a M (2;−1;0)∈d ; N (1;1;3)∈ d ⇒ MN (−1;2;3) yay 1 2  eaae MN (−1;2;3)           d1;d2 .MN A; B.C a e d =     ey (d1;d2 )     d1d; 2  A; B   aae A& B a eay A& B aa  yaya 11 y 3 a
  61. aeya(α ) ey         a ⊥ d1 ay a   (d1= A; d 2 = B) a⊥ d2       yaa a d1; d 2  a  a e y y a =( −1;2;2) (α ) a (α ) :−−+( x 2) 2( y ++ 1) 2( z) =⇔−+++= 0 x 2 y 2 z 3 0 yyay aaya yaa OE n n−1 aa P( x) = a0 x + a 1 x + + an ( x− c) a n−1 n − 2 Px( ) =( xcbx −)( 0 + bx 1 + + bxbn− 1 + n ) bi ( i = 0;1;2;3; ;n) e a a a a a a a a a P( x) =2 x4 + x 3 − 8 x 2 − x + 6 ae y P( x) =( x +2)( 2 x3 − 3 x 2 − 2 x + 3) + 0 yaaeaa yaaye aay 2x3 − 3( a + 1) x2 + 6 ax − 4 = 0( 1) 2x3 − 3( a + 1) x2 + 6 ax − 4 = 0 ay a a a 3 y ⇔( x −2)  2 x −( 3 a − 1) x + 2 = 0 ya mx3 −(3 m − 4) x2 +( 3 m − 7) x − m + 3 = 0 ( A)
  62. ToanDHSP.COM aaa ⇒ e 2 ( A) ⇔ ( x −1) mx − 2(m − 2) x + m − 3 = 0 ⇔ g ( x) = mx2 − 2(m − 2) x + m − 3 = 0 a  m ≠ 0  2 ∆ ' = (mm−m2) − ( − 3) > 0  m − 2 ⇔ S = > 0 ⇔∈m− ( ∞;0) ∪(3;4)  m  m − 3 P = > 0  m gm(1m) = − 2( − 2) + m − 3 ≠ 0 x3 −1− m( x −1) = 0 (1) (1) ⇔ x3 − mx + m −1 ya e y ⇔ ( x −1)( x2 + x +1− m) = 0 g(x) = x2 + x +1− m = 0 a  3 ∆ = 43m0− > m > 3 ⇔  ⇔  4 ⇔ < m ≠ 3 g (1)11=1 + + − m ≠ 0 4 m ≠ 3 ea y aa a x3 − 4mx2 + m2 x + 6m3 = 0 y (Q) = 2x5 + 3x4 − x3 − 7x2 +11x + 9 a aa aaa
  63. AO