怎么将y=x^-x/x^-x+1化简成(y-1)x^2+(1-y)x+y=0
1、x(x-y)(x+y)-x(x+y)^2
设x>1,y>0,若x^y+x^-y=2根号2,则x^y-x^-y等于
y=(x^2+x)/(x+1)
(1)(x^2/x)-y-x-y
若|x+y-1|+(x-y-2)²=0,求代数式(x+2y)(x-2y)-(2x-y)(-y-2x)的值.
先化简再求值(x-y)(x+y)-(x-2y) 的完全平方+x(3x-5y)-(x-y)(x-2y),其中x=1/2 y
若|x+2y-1|+y²+4y+4=0,求(2x-y)²-2(2x-y)(x+2y)+(x+2y)&
分式的加减:[(x/y-y/x)除以(x+y)+x(1/y-1/x)]除以1+x/y,其中x=5,y=2
将x(x+y)(x-y)-x(x+y)2进行因式分解,并求当x+y=1,xy=−12
(3x-y)^2+(3x+y)(3x-y),x=1,y=-2
{3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1
已知4x=9y求(1)x+y/y (2)y-x/2x