Question

化简求值[1-(2x/x+y）]÷ （x²-2xy+y²/3x+3y）+（x²+xy/x²-y²）其中x=-7，y=6

Answer 1

解：
1、化简
[1-(2x/x+y）]÷ （x²-2xy+y²/3x+3y）+（x²+xy/x²-y²）
=[(x+y)/(x+y)-(2x)/(x+y)]÷[(x-y)²/3(x+y)]+[x(x+y)/(x+y)(x-y)]
=[(x+y-2x)/(x+y)]÷[(x-y)²/3(x+y)]+x/(x-y)
=[-(x-y)/(x+y)][3(x+y)/(x-y)(x-y)]+x/(x-y)
=-3/(x-y)+x/(x-y)
=(x-3)/(x-y)
2、求值
∵
x=-7，y=6
∴
原式=(x-3)/(x-y)
=(-7-3)/(-7-6)
=(-10)/(-13)
=10/13

Answer 2

原题是不是这样的：[1-2x/（x+y）]÷ （x²-2xy+y²）/（3x+3y）+（x²+xy）/（x²-y²）其中x=-7，y=6
如果是=[(x+y)-2x]/(x+y)÷ (x-y)^2/[3(x+y)]+x(x+y)/[(x-y)(x+y)]
=-3(x-y)+x/(x-y)=(x-3)/(x-y)=(-7-3)/(-7-6)=10/13