Bài tập toán cao cấp part 4 pot

... ϕ.Tac´o x 4 + y 4 x 2 + y 2 = ρ 4 (cos 4 ϕ + sin 4 ϕ) ρ 2 (cos 2 ϕ + sin 2 ϕ) = ρ 2 (cos 4 ϕ + sin 4 ϕ). V`ı cos 4 ϕ + sin 4 ϕ  2nˆen lim x→0 y→0 x 4 + y 4 x 2 + y 2 = lim ρ→0 ρ 2 (cos 4 ϕ + sin 4 ϕ)=0. 7.3. ... x 0 nˆe ´ u: 1) f(x)=    4 · 3 x nˆe ´ u x<0 2a + x khi x  0. (DS. H`am f liˆen tu . c ∀x ∈ R nˆe ´ u a =2) 7 .4. Gi´o . iha . n v`a liˆen tu . ccu...

Ngày tải lên: 13/07/2014, 23:20

... lim x→0 1 − x 2 2 + x 4 8 + o(x 4 ) −1+ x 2 2 − x 4 24 +0(x 4 ) x 3 (x +0(x)) = lim x→0 x 4 8 − x 4 24 +0(x 4 ) x 4 +0(x 4 ) = lim x→0 1 8 − 1 24 + o(x 4 ) x 4 1+ 0(x 4 ) x 4 = 1 8 − 1 24 +0 1+0 = 1 12 · ... o(t 2 ) =  − x 2 2 + x 4 24 + o(x 4 ) − 1 2  − x 2 2 + x 4 4 + o(x 4 )  2  + o   − x 2 2 + x 4 24 + o(x 4 )  2  = − x 2 2 + x 4 24...

Ngày tải lên: 13/07/2014, 23:20

... a n = √ n 2 +4n 3 √ n 3 − 3n 2 .(DS. 1) 14. a n = (n + 3)! 2(n + 1)! − (n + 2)! .(DS. −∞) 15. a n = 2 +4+ ···+2n n +2 − 2. (DS. −1) 16. a n = n − 3 √ n 3 − n 2 .(DS. 1 3 ) 17. a n = 1 − 2+3− 4+ 5−···−2n √ n 2 +1+ √ 4n 2 +1 .(D S. ... 2 · 3 + 1 2 · 3 · 4 + ···+ 1 n(n + 1)(n +2) Chı ’ dˆa ˜ n. Tru . ´o . chˆe ´ ttach´u . ng minh r˘a ` ng 1 n(n + 1)(n +2) = 1 2  1 n(n +1) − 1 (n + 1)(n +2)...

Ngày tải lên: 13/07/2014, 23:20

... x 2 (DS. 1 2π ) 43 . lim x→0 cos mx − cos nx x 2 (DS. 1 2 (n 2 − m 2 )) 44 . lim x→∞ x 2  cos 1 x − cos 3 x  (DS. 4) 45 . lim x→0 sin(a + x) + sin(a −x) −2sin a x 2 (DS. −sin a) 46 . lim x→0 cos(a ... lim x→−2 x 2 − 4 arctg(x +2) (DS. 4) 38. lim x→0 tgx − sin x x 3 (DS. 1 2 ) 39. lim x→0 xcotg5x (DS. 1 5 ) 40 . lim x→1 (1 − x)tg πx 2 (DS. 2 π ) 41 . lim x→1 1 − x 2 sin πx (DS. 2 π...

Ngày tải lên: 13/07/2014, 23:20

... f(x)nˆe ´ u: 1. y = 4 √ x 3 + 5 x 2 − 3 x 3 + 2. (DS. 3 4 4 √ x − 10 x 3 + 9 x 4 ) 2. y = log 2 x + 3log 3 x.(DS. ln 24 xln2 · ln3 ) 3. y =5 x +6 x +  1 7  x .(DS. 5 x ln5 + 6 x ln6 − 7 −x ln7) 4. y = ln(x ... sin 4 x + cos 4 x.(DS. −sin 4x) 11. y = √ xe √ x .(DS. e √ x (1 + √ x) 2 √ x ) 12. y = e 1 cos x .(D S. e 1 cos x sin x cos 2 x ) 13. y = e 1 lnx .(DS. −e 1 lnx xln 2 x )...

Ngày tải lên: 13/07/2014, 23:20

... d[x √ 64 − x 2 ]+d  64arcsin x 8  = xd √ 64 − x 2 + √ 64 −x 2 dx +64d  arcsin x 8  = x d( 64 − x 2 ) 2 √ 64 − x 2 + √ 64 −x 2 dx + 64 d  x 8   1 − x 2 64 = −x 2 dx √ 64 −x 2 + √ 64 − x 2 dx ... f(x)=  ln 2 x − 4. d 2 f ?(DS. 4lnx 4 −ln 3 x x 2  (lnx − 4) 3 (dx) 2 ) 17. f(x) = sin 2 x. d 3 f ?(DS. 4 sin 2x(dx) 3 ) 18. f(x)= √ x −1, d 4 f ?(DS. −15 16(x −1) 7/2...

Ngày tải lên: 13/07/2014, 23:20

... y,x 2 + y 2 ). (D S. u  xx = f  tt +4xf  tv +4x 2 f  vv +2f  v , u  xy = f  tt +2(x + y)f  tv +4xyf  vv , u  yy = f  tt +4yf  tv +4y 2 f  vv +2f  v , t = x + y, v = x 2 + ... y 2 ∂ 2 u ∂y 2 = n(n −1)u. 44 . u = f(x + g( y)), ∂u ∂x · ∂ 2 u ∂x∂y = ∂u ∂y · ∂ 2 u ∂x 2 · 45 . T`ım da . o h`am theo hu . ´o . ng ϕ = 135 ◦ cu ’ a h`am sˆo ´ f(x, y)=3x 4 + xy + y 3 ta . i...

Ngày tải lên: 13/07/2014, 23:20

... ra  (1, 04) 1,99 + ln(1, 02) ≈ √ 1+ln1+1· (0, 04) + 0 · (−0, 01) +(1/2) · 0, 02=1,05. 4) Ta thˆa ´ y d l`a gi´a tri . cu ’ a h`am f(x, y, z)=2 x sin y arctgx ta . idiˆe ’ m M(−2, 95; 1, 49 ; 0, ... dy 2 ) + f  vv (0, 2)(dx + dy) 2 ) 24. u = f(x + y, z 2 ), M(x, y, z), M 0 (−1, −1, 0). (DS. d 2 u   M = f  tt (dx + dy) 2 +4zf  tv dz(dx + dy) +4z 2 f  vv dz 2 +2f  v d 2 z, d 2 u...

Ngày tải lên: 13/07/2014, 23:21

... f  x v`a f  y v`a t`ım c´ac diˆe ’ mt´o . iha . n. Ta c´o f  x =4x 3 − 4x +4y, f  y =4y 3 +4x −4y. Do d ´o 4x 3 − 4x +4y =0 4y 3 +4x −4y =0 158 Chu . o . ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe ` ubiˆe ´ n (DS. ... v´o . ic´acd iˆe ` ukiˆe . n x + y + z =5,xy + yz + zx =8. (D S. f max =4 4 27 ta . i  4 3 , 4 3 , 7 3  ;  4 3 , 7 3 , 4 3  ;  7 3 , 4 3 , 4 3  f min =4ta . i...

Ngày tải lên: 13/07/2014, 23:21