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[2/xy/(1/x+1/y)2+(x2+y2)/(x2+2xy+y2)]*2x/(x-y)

来源：学生作业帮 编辑：神马作文网作业帮 分类：数学作业 时间：2024/11/13 13:27:37
[2/xy/(1/x+1/y)2+(x2+y2)/(x2+2xy+y2)]*2x/(x-y)
[2/xy/(1/x+1/y)2+(x2+y2)/(x2+2xy+y2)]*2x/(x-y)
[2/xy/(1/x+1/y)²+(x²+y²)/(x²+2xy+y²)]*2x/(x-y)
=[2/xy/(x+y)²/x²y²+(x²+y²)/(x+y)²]*2x/(x-y)
=[2xy/(x+y)²+(x²+y²)/(x+y)²]*2x/(x-y)
=[(2xy+x²+y²)/(x+y)²]*2x/(x-y)
=(x+y)²/(x+y)²*2x/(x-y)
=2x/(x-y)
2x2+xy-3y2+x+4y-1因式分解 已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以（x2+y2/x 已知2x=3y,求xy／(x2+y2)-y2／(x2-y2)的值 已知x-y+1,X2+Y2=25 求（x+y）2和x2-xy+y2的值 x(x-1)-(x2-y)=-3,求x2-y2-2xy的值 x（x-1）-（x2-y）=-3,求x2+y2-2xy的值 已知x(x-1)-(x2-y)=-3,求x2+y2-2xy的值 当x=2,y=1 求{(x2+y2)/(x2-2xy+y2)+(2)/xy÷[(1/x)-(1/y)]}÷(x+y)的值 x2-xy-2y2-x+5y-2 x2-2xy+y2+3x-3y+2． x2-xy-2y2-x+5y-2因式分解 因式分解 x2+xy-2y2-x+7y-6