Question

-a(a^2-ab)除以b^2-2ab+a^2/ab乘以a-b/a^2

Answer 1

-a(a^2-ab)除以b^2-2ab+a^2/ab乘以a-b/a^2
=-a^2(a-b)*ab/(a-b)^2*(a-b)/a^2
=-ab

追问

解方程(1)1/y-2+3=y-1/y-2
(2)16/x^2-1=x+3/x-1

追答

(1)1/y-2+3=y-1/y-2
1+3(y-2)=y-1 y=1
经检验 y=1
是原方程的解
(2)16/x^2-1=x+3/x-1
16=(x+3)(x+1) x^2+4x-12=0 x=2, -6
经检验x=2, -6都是原方程的解

追问

先化简a^2-2a/a^2-1除以(a-1-2a-1/a+1)其中a=1/2012这个怎么做

追答

a-1-2a-1/a+1=(a^2-2a)/(a+1)
a^2-2a/a^2-1除以(a-1-2a-1/a+1)
=a^2-2a/a^2-1*(a^2-2a)/(a+1)
=1/(a-1)
=1/(1/2012-1)
=-2012/2011

追问

当x=1+∫3时求式子(x^2-1)(x+1)/x^2-1除以(1+x^2+1/2x)的值

追答

(x^2-1)(x+1)/x^2-1除以(1+x^2+1/2x)
=(x+1)*2x/(x+1)^2=2x/(x+1)=2(1+∫3)/(2+∫3)=2(1+∫3)/(2-∫3)=2∫3-2