①(2y-z)²[2y(z+2y)+z²]²=
①(2y-z)²[2y(z+2y)+z²]²=
若x-y=6,xy=-8,求代数式(x+y+z)²+(x-y-z)(x-y+z)-2·z(x+y)的值
(x-3y+2z)²=
已知x²+y²+z²-2x+4y-6z+14=0,求x+y-z/x+y+z的值
已知x,y,z满足x+y+2z=1,x²+y²+6z+1.5=0,求x,y,z的值
3(y-z)²-(2y+z)(-y+2y)
已知x-y=2,y-z=2,x+z=14,求x²-z²的值
已知x-y=2,y-z=2,x+z=14,求x²-z²的值.
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z
若3x-2y-z=0,x+4y-3z=0,求3x²+y²/4x²-y²+z
已知实数x,y,z,满足x²+4y²+根号-z²=2x+4y-2,求x+2y-z的平方