初二数学问题:(x+2y-z)(x-2y+z)²-(x+2y+z)² 计算过程
已知X/2=y/3=z/4,试求x+y-z/x+y+z的值 (这是初二分式) 要过程
初中数学整式加减问题已知x-y=2,y+z=3,求多项式(x-y)²+(x+z)²-(y+z)&su
x,y,z为正实数 x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)
运用乘法公式计算:(x+2y-3z)(x-2y+3z)
(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z
①(x+y+z)(-x+y+z)(x-y+z)(x+y-z)
计算:(x+y+z)(-x+y+z)(x-y+z)(x+y-z).
{ x+y+z=2 x-z=-2 y+z=1的解(求过程)
x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x-2z+y)(y+z-2x)+(x
证明:(y+z-2x)3+(z+x-2y)3+(x+y-2z)3=3(y+z-2x)(z+x-2y)(x+y-2z).
计算:(x-2y+z)(-x+2y+z)要过程,谢