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

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

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

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
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