x+y+z=26 x-y=1 2x-y+z=18
(x+y-z)(x-y+z)=
已知 x,y,z都是正实数,且 x+y+z=xyz 证明 (y+x)/z+(y+z)/x+(z+x)/y≥2(1/x+1
解方程组;1.x+y+z=26,x-y=1,2x-y+z=18 2.5x+y+z=1,2x-y+2z=1,x+5y-z=
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
X+Y+Z=?
f(x,y,z,w)=x*(x+y)*(x+y+z)*(x+y+z+w)
(x+y-z)^2-(x-y+z)^2=?
x,y,z为实数 且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+z-2y)^2+(x+y
x,y,z为实数且(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y-
解方程组{x(x+y+z)=6,y(x+y+z)=12,z(x+y+z)=18
分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y)
解三元一次方程 X+Y+Z=26 X-Y=1 2X-Y+Z=18