y=(x³-2)/[2(x-1)²]，求y'解：y'=[(x-1)²(3x²)-2(x³-2)(x-1)]/[2(x-1)⁴]=[3x²(x-1)-2(x³-2)]/[2(x-1)³]=(x³-3x²+4)/[2(x-1)³]
应该是吧上面的变成x3-1-1利用立方差公式变成两个相减吧
(2-x^2)/((x-1)^3)
