(16x^4+y^4)(4x^2+2y^2)(2x-y)(2x+y)的答案是什么?
{(z+y)(x-y)-(x-y)的2次方+2y(x-y)}除以4y
x^2-y^2/x+y-4x(x-y)+y^2/2x-y,
(2x-y)^2-4(x-y)(x+2y)
(2x-y)²-4(x-y)(x+2y)
化简:(2x-y)²-4(x-y)(x+y)
x+2y+4y^2/x-y+4x^y/4y^2-x^2
4x(y-x)-y^2因式分解
(x+3y-4)(2x-y)
已知x*x-4xy+4y*y=0 求[2x(x+y)-y(x+y)]/(4x*x-4xy+y*y)的值?
x/x-y*y^2/x+y-x^3y2/x^4-y^4/x^2/x^2+y^2
(x-2y)(x-2y)(x平方+4y平方)的平方(x+2y)(x+2y)
(X-3Y)(X+3Y)-(X+2Y)(X-4Y)-(2X-Y)2的平方