作业帮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-](/uploads/image/z/4296554-26-4.jpg?t=f%28x%29%3D%EF%BC%88x%5E3-2%EF%BC%89%2F%5B2%28x-1%29%5E2%5D%E6%B1%82%E5%AF%BCf%27%28x%29%3D%5B6x%26%23178%3B%28x-1%29%26%23178%3B-4%28x%26%23179%3B-2%29%28x-1%29%5D%2F%5B2%28x-1%29%26%23178%3B%5D%26%23178%3B%3E0%E5%8D%B3+6x%26%23178%3B%28x-1%29%26%23178%3B-4%28x%26%23179%3B-2%29%28x-1%29%3E0+%28x-1%29%5B6x%26%23178%3B%28x-1%29-4%28x%26%23179%3B-2%29%5D%3E0+%28x-1%29%5B6x%26%23179%3B-6x%26%23178%3B-4x%26%23179%3B%2B8%5D%3E0+%28x-1%29%28x%26%23179%3B-3x%26%23178%3B%2B4%29%3E0+%28x-)
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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