:(2x-y)(2x-y)=-5/2(2y^2-64/5),2x(y-1)+4(1/2x-1)=0,求下列各式的值(x-y)^2(2)x^4+y^4-2x^2y^2已知:(2x-y)(2x-y)=-5/2(2y^2-64/5),2x(y-1)+4(1/2x-1)=0,求下列各式的值（x-y）^2(2)x^4+y^4-2x^2y^2
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:(2x-y)(2x-y)=-5/2(2y^2-64/5),2x(y-1)+4(1/2x-1)=0,求下列各式的值(x-y)^2(2)x^4+y^4-2x^2y^2已知:(2x-y)(2x-y)=-5
:(2x-y)(2x-y)=-5/2(2y^2-64/5),2x(y-1)+4(1/2x-1)=0,求下列各式的值(x-y)^2(2)x^4+y^4-2x^2y^2已知:(2x-y)(2x-y)=-5/2(2y^2-64/5),2x(y-1)+4(1/2x-1)=0,求下列各式的值（x-y）^2(2)x^4+y^4-2x^2y^2
:(2x-y)(2x-y)=-5/2(2y^2-64/5),2x(y-1)+4(1/2x-1)=0,求下列各式的值(x-y)^2(2)x^4+y^4-2x^2y^2
已知:(2x-y)(2x-y)=-5/2(2y^2-64/5),2x(y-1)+4(1/2x-1)=0,求下列各式的值（x-y）^2(2)x^4+y^4-2x^2y^2
:(2x-y)(2x-y)=-5/2(2y^2-64/5),2x(y-1)+4(1/2x-1)=0,求下列各式的值(x-y)^2(2)x^4+y^4-2x^2y^2已知:(2x-y)(2x-y)=-5/2(2y^2-64/5),2x(y-1)+4(1/2x-1)=0,求下列各式的值（x-y）^2(2)x^4+y^4-2x^2y^2
条件书写可能有误且不够明确,看了让人觉得不好回答；
如条件开头是 (2x-y)(2x+y)=……,则：
由 (2x-y)(2x+y)=-(5/2)(2y^2-64/5) 得：4x²-y²=5y²+32,即 x²+y²=8；
由 2x(y-1)+4[(x/2)-1]=0 得：2xy-2x+2x-4=0,即 xy=2；
∴ (x-y)²=x²+y²-2xy=8-2*2=4；
x^4+y^4=(x²+y²)²-4(xy)²=8²-4*2²=48；
x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2
已知x-y/x+y=3,求代数式5(x-y)/x+y-x+y/2(x-y)
2(x-y)=5y,x-y=-5
x+y/2+x-y/6=5 2(x+y)-3(x-y)=-2
x-y / 5y= (y-x)^2 / ( )
解方程组(x+y)(x-2y)=-5 (x-y)(x+2y)=7
3(x+y)+2(x-y)=36 4(x+y)-5(x-y)=2
x*x+2x+y*y-4y+5=0 x= y=
4(x-y)-3(x-y)=9 5(x-y)+2(x-y)=11
3(x+y)-4(x-y)=11,2(x-y)+5(x+y)=27
{6(x-y)-7(x+y)=21 {2(x-y)-5(x+y)=-1
6(x-y)-2(x+y)=14 3(x-y)+(x+y)=5
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{(x+y)/2+(x-y)/3=1,(x+y)-5(x-y)=2.求x=?,y=?