f(x)=(x^3-2)/[2(x-1)^2]求导f'(x)=[6x²(x-1)²-4(x³-2)(x-1)]/[2(x-1)²]²>0即 6x²(x-1)²-4(x³-2)(x-1)>0 (x-1)[6x²(x-1)-4(x³-2)]>0 (x-1)[6x³-6x²-4x³+8]>0 (x-1)(x³-3x²+4)>0 (x-

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f(x)=(x^3-2)/[2(x-1)^2]求导f'(x)=[6x²(x-1)²-4(x³-2)(x-1)]/[2(x-1)²]²>0即 6x²(x-1)²-4(x³-2)(x-1)>0 (x-1)[6x²(x-1)-4(x³-2)]>0 (x-1)[6x³-6x²-4x³+8]>0 (x-1)(x³-3x²+4)>0 (x-

f(x)=(x^3-2)/[2(x-1)^2]求导f'(x)=[6x²(x-1)²-4(x³-2)(x-1)]/[2(x-1)²]²>0即 6x²(x-1)²-4(x³-2)(x-1)>0 (x-1)[6x²(x-1)-4(x³-2)]>0 (x-1)[6x³-6x²-4x³+8]>0 (x-1)(x³-3x²+4)>0 (x-
f(x)=(x^3-2)/[2(x-1)^2]求导
f'(x)=[6x²(x-1)²-4(x³-2)(x-1)]/[2(x-1)²]²>0即 6x²(x-1)²-4(x³-2)(x-1)>0 (x-1)[6x²(x-1)-4(x³-2)]>0 (x-1)[6x³-6x²-4x³+8]>0 (x-1)(x³-3x²+4)>0 (x-1)[x³+x²-4x²+4]>0 (x-1)[x²(x+1)-4(x-1)(x+1)]>0 (x-1)(x+1)(x²-4+4)>0 (x-1)>0
我想问从(x-1)(x³-3x²+4)>0想到 (x-1)[x³+x²-4x²+4]>0到底是怎么想出来的?
用了什么思想?或者说x^3-3x^2+4x怎么化成(x+1)(x-2)²

f(x)=(x^3-2)/[2(x-1)^2]求导f'(x)=[6x²(x-1)²-4(x³-2)(x-1)]/[2(x-1)²]²>0即 6x²(x-1)²-4(x³-2)(x-1)>0 (x-1)[6x²(x-1)-4(x³-2)]>0 (x-1)[6x³-6x²-4x³+8]>0 (x-1)(x³-3x²+4)>0 (x-
显然-1是x³-3x²+4=0的一个根,用大除法就能快速解决了:
x³-3x²+4=(x+1)(x2-4x+4)=(x+1)(x-2)²
你那个过程纯粹是在秀技巧,没什么价值

不vnvbnfdgkdfjgkljdfklgjdfg

啦啦啦 表示不会= =

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