题目

计算："3y/x+y"+"y/y-x"+"2x/x^2-y^2" "a+2/a^2-4a+4"-"1/2a+4"-"a-1/2a^2-8"
a^2+a+1-"a^3/a-1" "1/a-2"-"12/a^2-4"
一共4题。

提问时间：2020-08-01

答案

"3y/x+y"+"y/y-x"+"2x/x^2-y^2"
=[3y*(x-y)-y(x+y)]/[(x+y)(x-Y)]
=(2xy-4y^2)/[(x-y)(x+y)]
=2xy/(x+y)
a+2/a^2-4a+4"-"1/2a+4"-"a-1/2a^2-8
=(a+2)/(a-2)^2-1/[2(a+2)]-(a-1)/[2(a+2)(a-2)]
=(a+2)/(a-2)^2-(a-2)/[2(a+2)(a-2)]-(a-1)/[2(a+2)(a-2)]
=(a+2)/(a-2)^2-[(2a-3)/[2(a+2)(a-2)]
=[2*(a+2)^2-(2a-3)(a-2)]/[2(a+2)(a-2)^2]
=(15a+2)/2(a-2)^2(a+2)
3.a^2+a+1-"a^3/a-1
=[(a^2+a+1)(a-1)-a^3]/(a-1)
=-1/(a-1)
4.1/a-2"-"12/a^2-4
=1/(a-2)-12/(a+2)(a-2)
=[(a+2)-12]/(a+2)(a-2)
=(a-10)/(a+2)(a-2)