初一数学代数式求值，已知1/x+1/y=2，求(2x+5xy+2y)/(-x+3xy-y)
\u521d\u4e00\u6570\u5b66 \u5df2\u77e5x^4+y^4=50,x^2y-xy^2=14,\u6c42\u4ee3\u6570\u5f0fx^4-y^4+3xy^2-5x^2y+3x^2y+2y^4\u7684\u503cx^4- y^4+3xy^2-xy^2-5x^2y+3x^2y+2y^4
=x^4+ y^4+2xy^2-2x^2y
=50-2*(-14)
=78
\u7531y=3xy+x\u53ef\u5f97\uff1ax-y=-3xy
2x+3xy-2y/x-2xy-y
=2\uff08x-y\uff09+3xy/(x-y)-2xy
=(-6xy+3xy)/(-3xy-2xy)
=-3xy/-5xy
=3/5
由(2x+5xy+2y)/(-x+3xy-y)分子分母分别合并得[2（x+y）+5xy]/[-(x+y)+3xy]②
把①代入②得(4xy+5xy)/(-2xy+3xy)=9xy/xy=9
希望你满意......
x+y=2xy,原式化为：9xy/xy=9
1/X+1/Y=2 左右都乘以XY 则 X+Y=2XY X-2XY+Y=0
原式化简为2（X+5/2XY+Y）/-（X-3XY+Y）=2[(X-2XY+Y)+9/2XY]/-[(X-2XY+Y)-XY]
=2(9/2XY)/-(-XY)=9XY/XY=9
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