các dạng bài tập toán cao cấp 3

...                                 = =        − = =        = − =        − = − = ⇔                    = − =       = − = ⇔      =− =− ⇔    = = 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 0 13 0 13 0' 0' 2 2 y x y x y x y x y y x x y x z z y x ⇒h/số có 4 điểm tới hạn: ) 3 1 , 3 1 ( 4 ), 3 1 , 3 1 ( 3 ) 3 1 , 3 1 ( 2 ), 3 1 , 3 1 ( 1 MM MM − −−− Ta ... 0154.41 4 4 1'' 1'' 4 4 1'' ) 3 32 , 3 32 ( 2 3 4 )( )( 3 4 )( <−=−=−⇒ =+== == =+== ⇒ rts zt zs zr M Myy Mxy Mxx và r = 4 > 0 ⇒ h/số cực tiểu tại: ) 3 32 , 3 32 (M và 3 4 ln74 3 4 ln 2 14 3 4 .3) 3 32 , 3 32 ( min −=−== zZ Câu ... ] xCx CxeZ x x dx dxx x dxxe dxxQeJ xdx x dxPI CdxQe eZ x x dxP x x dxP dxP x x x ln3 ln3 ln 33 )3. ( 1 )3. ( )(. ln3 3 . 3 ln3 2 3 2ln3 )( )( )( )( )( = += == = = = = == + ì = Thay: [ ] xCx yy Z ln3 11 3 33 −=⇒= ⇒No t/quát: [ ] xCx y ln3 3 3 1 −= (C là hằngsố) b....

... sin 5π 6 ) 6) 1 2 − i √ 3 2 (D S. cos 5π 3 + i sin 5π 3 ) 7) − 1 2 − i √ 3 2 (D S. cos 4π 3 + i sin 4π 3 ) 8) 2 + √ 3 − i (DS. 2  2+ √ 3  cos 23 12 + i sin 23 12  ) 9) 2 − √ 3 − i (DS. 2  2 − √ 3  cos 19π 12 + ... =  −1+i √ 3 2  3  m +  −1 − i √ 3 2  3  m =  −1+3i √ 3+ 9− 3i √ 3 8  m +  −1 − 3i √ 3+ 9+3i √ 3 8  m =1 m +1 m =2. 1.4. Biˆe ’ udiˆe ˜ nsˆo ´ ph´u . cdu . ´o . ida . ng lu . o . . ng gi´ac 35 Gia ’ i. 1) ... − z −1 2i  5 = z 5 − 5z 3 +10z − 10z −1 +5z 3 − z −5 32 i = (z 5 − z −5 ) − 5(z 3 − z 3 )+10(z − z −1 ) 32 i = 2i sin 5ϕ − 10i sin 3 +20i sin ϕ 32 i = sin 5ϕ− 5 sin 3 + 10 sin ϕ 16 · Chu . o . ng...

... sˆo ´ : 0, 2; 0, 23; 0, 233 ; 0, 233 3; ,0, 233 3    n , Chı ’ dˆa ˜ n. Biˆe ’ udiˆe ˜ n a n du . ´o . ida . ng a n = 2 10 +  3 10 2 + 3 10 3 + ···+ 3 10 n  (D S. 7 /30 ) 13. Ch´u . ng minh ... c´o: a n =  3 √ n +2  3 −  3 √ n  3  3 √ n +2  2 + 3 √ n +2· 3 √ n +  3 √ n  2 a n = 2  3 √ n +2  2 + 3 √ n +2· 3 √ n +  3 √ n  2 Biˆe ’ uth´u . cmˆa ˜ usˆo ´ b˘a ` ng: n 2 /3  3  1+2/n  2 + 3  1+2/n ... n 3 + n  3 √ n 2 − n 3  2 − n 3 √ n 2 − n 3 + n 2   3 √ n 2 − n 3  2 − n 3 √ n 2 − n 3 + n 2 = n 2  3 √ n 2 − n 3  2 − n 3 √ n 2 − n 3 + n 2 = 1 [1/n − 1] 2 /3 − [1/n − 1] 1 /3 +1 suy ra...

... so . cˆa ´ p 43 T`u . d ´o x 3 = 1 t 3 − 1 , 1+x 3 = t 3 t 3 − 1 ,x=(t 3 − 1) − 1 3 dx = −t 2 (t 3 − 1) − 4 3 dt, x −2 =(t 3 − 1) 2 3 . Do vˆa . y I 3 = −  (t 3 − 1) 2 /3  t 3 t 3 − 1  −5 /3 t 2 (t 3 − ... − 1 8 x −1 (3x + 4)(2 + x 3 ) − 2 3 ) 30 .  dx √ x 3 3  1+ 4 √ x 3 .(DS. −2 3  (x − 3 4 +1) 2 ) 31 .  dx 3 √ x 2 ( 3 √ x +1) 3 .(DS. − 3 2( 3 √ x +1) 2 ) 32 .  3 √ x  3 √ x +1 dx. (D S. 6  u 7 7 − 3 5 u 5 + ... cos 3xdx = − 1 3 x 2 cos 3x + 2 3 I 1 . Ta cˆa ` n t´ınh I 1 .D˘a . t u = x, dv = cos 3xdx. Khi d´o du =1dx, v = 1 3 sin 3x.T`u . d ´o I = − 1 3 x 2 cos 3x + 2 3  1 3 x sin 3x − 1 3  sin 3xdx  =...

... sˆo ´ : 0, 2; 0, 23; 0, 233 ; 0, 233 3; ,0, 233 3    n , Chı ’ dˆa ˜ n. Biˆe ’ udiˆe ˜ n a n du . ´o . ida . ng a n = 2 10 +  3 10 2 + 3 10 3 + ···+ 3 10 n  (D S. 7 /30 ) 13. Ch´u . ng minh ... n 3 + n  3 √ n 2 − n 3  2 − n 3 √ n 2 − n 3 + n 2   3 √ n 2 − n 3  2 − n 3 √ n 2 − n 3 + n 2 = n 2  3 √ n 2 − n 3  2 − n 3 √ n 2 − n 3 + n 2 = 1 [1/n − 1] 2 /3 − [1/n − 1] 1 /3 +1 suy ra ... c´o: a n =  3 √ n +2  3 −  3 √ n  3  3 √ n +2  2 + 3 √ n +2· 3 √ n +  3 √ n  2 a n = 2  3 √ n +2  2 + 3 √ n +2· 3 √ n +  3 √ n  2 Biˆe ’ uth´u . cmˆa ˜ usˆo ´ b˘a ` ng: n 2 /3  3  1+2/n  2 + 3  1+2/n...

... 1) 2 n , 1+ 1 3 + ···+ 1 3 n = 3( 3 n − 1) 2 · 3 n v`a do d´o: lim a n = lim 2(2 n − 1) 2 n · 2 · 3 n 3( 3 n − 1) = 2 lim 2 n − 1 2 n · 2 3 lim 3 n 3 n − 1 = 2 lim[1 − (1/2) n ] · 2 3 lim 1 1 − (1 /3) n =2· ... =    cos(n +1) 3 n+1 + ···+ cos(n + p) 3 n+p     1 3 n+1 + ···+ 1 3 n+p = 1 3 n+1 1 − 1 3 p 1 − 1 3 < 1 2 · 1 3 n < 1 3 n · Gia ’ su . ’ ε l`a sˆo ´ du . o . ng t`uy ´y. V`ı lim n→∞ 1 3 n = ... lim[1 − (1/2) n ] · 2 3 lim 1 1 − (1 /3) n =2· 1 · 2 3 · 1= 4 3 ·  V´ı d u . 3. 1) a n = √ n 2 + n − n 2) a n = 3 √ n +2− 3 √ n 3) a n = 3 √ n 2 − n 3 + n Gia ’ i. 1) Ta biˆe ´ nd ˆo ’ i a n b˘a ` ng...

... 3 ∀u ∈ X. T ` ’ u (i) v`a (ii), ta suy ra y, u = Ax, u ⇔ y − Ax, u = 0, ∀u ∈ X ⇔ y − Ax = ... ⇔ y = Ax. Do ¯d´o (x, y) = (x, Ax) ∈ G(A). * V`ı X l`a khˆong gian Banach nˆen A liˆen tu . c. 13. Gi ’ a s ’ ’ u {e n } n l`a mˆo . t c ’ o s ’ ’ o c ’ ua khˆong gian Hilbert X v`a P n x = n  k=1 x,...

... sˆo ´ : 0, 2; 0, 23; 0, 233 ; 0, 233 3; ,0, 233 3    n , Chı ’ dˆa ˜ n. Biˆe ’ udiˆe ˜ n a n du . ´o . ida . ng a n = 2 10 +  3 10 2 + 3 10 3 + ···+ 3 10 n  (D S. 7 /30 ) 13. Ch´u . ng minh ... a n =0. 3) Ta c´o thˆe ’ viˆe ´ t n = 3 √ n 3 v`a ´ap du . ng cˆong th´u . c: a 3 + b 3 =(a + b)(a 2 − ab + b 2 ) suy ra a n =  3 √ n 2 − n 3 + n  3 √ n 2 − n 3  2 − n 3 √ n 2 − n 3 + n 2   3 √ n 2 − ... bˆa ´ td˘a ’ ng th´u . c (*) ta Bài tập toán cao cấp Tập 2 Nguyễn Thủy Thanh NXB Đại học quốc gia Hà Nội 2007, 158 Tr. Từ khoá: Bài tập toán cao cấp, Giới hạn dãy số, Giới hạn hàm...

... 1 + Nˆe ´ u n =3m th`ı S =  −1+i √ 3 2  3  m +  −1 − i √ 3 2  3  m =  −1+3i √ 3+ 9− 3i √ 3 8  m +  −1 − 3i √ 3+ 9+3i √ 3 8  m =1 m +1 m =2. 1.1. D - i . nh ngh˜ıa sˆo ´ ph´u . c 7 (a 1 ,b 1 )+(a 2 ,b 2 ) def =(a 1 + ... bˆo . icu ’ a3th`ı  −1+i √ 3 2  n +  −1 − i √ 3 2  n =2 v`a nˆe ´ u n khˆong chia hˆe ´ t cho 3 th`ı  −1+i √ 3 2  n +  −1 − i √ 3 2  n = −1. Gia ’ i. 1 + Nˆe ´ u n =3m th`ı S =  −1+i √ 3 2  3  m +  −1 ... − z −1 2i  5 = z 5 − 5z 3 +10z − 10z −1 +5z 3 − z −5 32 i = (z 5 − z −5 ) − 5(z 3 − z 3 )+10(z − z −1 ) 32 i = 2i sin 5ϕ − 10i sin 3 +20i sin ϕ 32 i = sin 5ϕ − 5 sin 3 + 10 sin ϕ 16 · 1.2....

... có:             − − − −− → → → →             − − − −− ↔             − − − −− → → → →             − − − −− →+ →+ →+ →             −− −− − −− = 2221000 701100 31 0 033 0 712011 701100 2221000 31 0 033 0 712011 701010 2221000 31 0 033 0 712101 701010 31 0 033 0 2221000 712101 3 2 011111 1 036 033 8 032 02 712101 43 34 22 11 32 44 32 23 11 441 33 1 221 11 hh hh hh hh cc hh hh hh hh hhh hhh hhh hh A ⇒ r(A) ...             −− → → →           −− →+ → →           −− −− ↔+ →+ →−           −−− − 3 1 3 1 3 1 100 012710 001201 6 1 222600 012710 001201 210110 012710 001201 2 2 100211 01 031 2 001201 33 22 11 33 2 22 11 32 3 221 11 hh hh hh hhh hh hh hhh hhh hh                 −−− − → → →+                 −−− −− → →+− → 3 1 3 1 3 1 100 3 7 3 4 3 1 010 3 2 3 2 3 1 001 2 3 1 3 1 3 1 100 3 7 3 4 3 1 010 001201 7 33 22 1 13 33 2 23 11 hh hh hhh hh hhh hh Ta có:           − − −− = 232 110 121 t A ;           −−= 011 212 030 t B ;           −− − − = 244 285 012 t AB ;                 − −− − = − 3 2 3 1 3 1 3 5 3 16 3 4 3 1 3 8 3 6 1 BA Vậy                 −− − −− =+++ − 3 8 3 4 3 5 3 14 3 49 3 11 3 2 3 5 5 2 1 ... Ngũ                 − −− → →+− →               − − → → →+           − − − →+ → →           − − − → →+− →           − − 6 1 6 1 6 3 100 6 5 6 1 6 3 010 6 1 6 1 6 3 001 5 6 1 6 1 6 3 100 0 135 10 6 1 6 1 6 3 001 6 1 6 1 1 136 00 0 135 10 001101 100110 0 135 10 001101 3 100110 0102 13 001101 33 2 23 11 33 22 1 13 332 22 11 33 221 11 hh hhh hh hh hh hhh hhh hh hh hh hhh hh Ta...

... sin 5π 6 ) 6) 1 2 − i √ 3 2 (D S. cos 5π 3 + i sin 5π 3 ) 7) − 1 2 − i √ 3 2 (D S. cos 4π 3 + i sin 4π 3 ) 8) 2 + √ 3 − i (DS. 2  2+ √ 3  cos 23 12 + i sin 23 12  ) 9) 2 − √ 3 − i (DS. 2  2 − √ 3  cos 19π 12 + ... Sˆo ´ ph´u . c 2 + Nˆe ´ u n =3m +1th`ı S =  −1+i √ 3 2  3  m  −1+i √ 3 2  +  −1 − i √ 3 2  3  m  1 − i √ 3 2  = −1+i √ 3 2 + −1 − i √ 3 2 = −1. Tu . o . ng tu . . nˆe ´ u n =3m +2tac˜ung c´o ... = (− √ 3+ i)  cos π 12 − i sin π 12  1 − i · 2) z =  √ 3+ i. 2MU . CLU . C 3. 2.4 Phu . o . ng ph´ap t´ınh d i . nh th´u . c 89 3. 3 Ha . ng cu ’ a ma trˆa . n 109 3. 3.1 D - i . nhngh˜ıa 109 3. 3.2...

... 1 + Nˆe ´ u n =3m th`ı S =  −1+i √ 3 2  3  m +  −1 − i √ 3 2  3  m =  −1+3i √ 3+ 9− 3i √ 3 8  m +  −1 − 3i √ 3+ 9+3i √ 3 8  m =1 m +1 m =2. 26 Chu . o . ng 1. Sˆo ´ ph´u . c 2) Trong ... bˆo . icu ’ a3th`ı  −1+i √ 3 2  n +  −1 − i √ 3 2  n =2 v`a nˆe ´ u n khˆong chia hˆe ´ t cho 3 th`ı  −1+i √ 3 2  n +  −1 − i √ 3 2  n = −1. Gia ’ i. 1 + Nˆe ´ u n =3m th`ı S =  −1+i √ 3 2  3  m +  −1 ... − z −1 2i  5 = z 5 − 5z 3 +10z − 10z −1 +5z 3 − z −5 32 i = (z 5 − z −5 ) − 5(z 3 − z 3 )+10(z − z −1 ) 32 i = 2i sin 5ϕ − 10i sin 3 +20i sin ϕ 32 i = sin 5ϕ− 5 sin 3 + 10 sin ϕ 16 · 1.2....

... sˆo ´ : 0, 2; 0, 23; 0, 233 ; 0, 233 3; ,0, 233 3    n , Chı ’ dˆa ˜ n. Biˆe ’ udiˆe ˜ n a n du . ´o . ida . ng a n = 2 10 +  3 10 2 + 3 10 3 + ···+ 3 10 n  (D S. 7 /30 ) 13. Ch´u . ng minh ... =    cos(n +1) 3 n+1 + ···+ cos(n + p) 3 n+p     1 3 n+1 + ···+ 1 3 n+p = 1 3 n+1 1 − 1 3 p 1 − 1 3 < 1 2 · 1 3 n < 1 3 n · Gia ’ su . ’ ε l`a sˆo ´ du . o . ng t`uy ´y. V`ı lim n→∞ 1 3 n = ... n.(DS. 0) 13. a n = √ n 2 +4n 3 √ n 3 − 3n 2 .(DS. 1) 14. a n = (n + 3) ! 2(n + 1)! − (n + 2)! .(D S. −∞) 15. a n = 2+4+···+2n n +2 − 2. (D S. −1) 16. a n = n − 3 √ n 3 − n 2 .(DS. 1 3 ) 17. a n = 1...

... = ĐS: 3 2 a π 30 . 2 2 2 , 2z x y z= + = ĐS: 4 π 31 . 2 2 2 2 2 , 2z x y x y z= + + + = ĐS: [8 2-7] 6 π 32 . 2 2 2 2 ,z x y z x y= + = + ĐS: 6 π 33 . 2 2 2 2 2 2 1 x y z a b c + + = ĐS: 4 3 abc π 10 ... = ĐS: ( ) 4 2 2 1 3 − 70.Phần mặt cầu 2 2 2 2 x y z R+ + = nằm trong mặt trụ 2 2 2 ,x y a a R+ = ≤ ĐS: ( ) 2 2 4 a a a R π − − Bài tập phần tích phân 3 lớp Bài 1: Tính các tích phân lặp sau: ... z= + + = = ĐS: 3 2 3 a π 57. 2 2 2 2 2 , , 0z x y x y a z= + + = = ĐS: 4 2 a π 58. 2 2 2 , , 0z x x y a z= + = = ĐS: 3 4 3 a 59. 2 2 4 , 1, 1z x y x y= − − = ± = ± ĐS: 1 13 3 60. 2 2 2 0,...

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