2y² - y = 1
代数式4y² - 2y + 5 = 7,那么代数式2y² - y + 1
若|x+2y-1|+y²+4y+4=0,求(2x-y)²-2(2x-y)(x+2y)+(x+2y)&
先化简,在求值:6y-(y+2分之y + 2-y分之y)除以y³-4y-2y²+8分之y,其中y=
解方程1/1-y²=2/1+2y+y²-3/1-2y+y²
[(y-2x)(-2x-y)-4(x-2y)²]*2y,其中x=1,y=2
解分式方程 2/y-y/(y-2)=4/(y²-2y)
解方程:3y^2-(2y+1)(y-1)=(y-5)(y+1)
解方程3y^2-(2y+1)(y-2)=(y-5)(y-1)
已知{(x²+y²)-(x-y)²+2y(x-y)}÷4y=1,求4x²-y
已知x²+y²+5=2x+4y,求【2x²-(x-y)(x-y)】【(x+y-1)(x-y
[2x²-(x+y) (x-y)] [(-x-y) (y-x)+2y²] 其中x=1,y=2
化简求值 【2x²-(x+y)(x-y)][(x+y-1)(x-y+1)+1-2y],其中x=-1,y=-2