- 急！初二数学题明天报名后天考试...急啊！！！一计算[x
- 明天报名后天...急啊！！！
一计算[x-1/（x^2-x-2）]^2/(x^2-2x+1)/(2-x)/[(1/x^2+x)]^2.
二先化简，再求值[-xy/(x-y）^2]^4*[(x^2-xy)/x]^3*x^4/y^10/[x/(xy-y^2)]^5,其中x=-2,y=4.
- 1.原式={x^4/[(x^2+1)(x+1）]}^2/(x-1)^2/(2-x)*x^2(x+1)^2
=x^8*x^2/(x^2+1)^2*(2-x)/(x-1)^2
=(2-x)x^10/[(x^2+1)(x-1)^2].
2.原式=x^4y^4/(x-y)^8*(x-y)^3*x^4/y^10*(x-y)^5y^5/x^5
=x^3/y,
当x=-2,y=4时原式=（-2）^3/4=-2.