几题分式计算（1）2/(m-1)^2除以（1-m）（2）(3x-6)/(x^2-4)除以(x+2)/(x^2+4x+4)

（1）
[2/(m-1)²] ÷ (1-m)
= [2/(1-m)²] ÷ (1-m)
= 2/(1-m)³
（2）
[(3x-6)/(x²-4)] ÷ [(x+2)/(x²+4x+4)]
= {2(x-2)/[(x+2)(x-2)]} ÷ [(x+2)/(x+2)²]
= [2/(x+2)] ÷ [1/(x+2)]
= 2
（3）
[(-x/y)²×(-y²/x)³] ÷ [(-xy)^4]
= [(x²/y²)×(-y^6)/x³] ÷ [(xy)^4]
= [(-y^4)/x] ÷ [(x^4)(y^4)]
= -1/x^5
（4） 此题似乎少加了些括号,以下是加了括号的一种形式
{[(2x-6)/(-5x²)]²×[(x+3)/(x²-6x+9)]} ÷ [(x+3)/(x-3)]²
= {[4(x-3)²/(-5x²)²]×[(x+3)/(x-3)²]} ÷ [(x+3)²/(x-3)²]
= [4(x+3)/(25x^4)] ÷ [(x+3)²/(x-3)²]
= 4(x-3)²/[(25x^4)(x+3)]
= (4x²-24x+36)/(25x^5+75x^4)

这些题目只要按照分式计算的规则一步步做就好