2x+y+z=15 x+2y+z=16 x+y+2z=17
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
解方程组:2x+y+z=15 x+2y+z=16 x+y+z=17
(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
分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^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-
x/2=y/3=z/5 x+3y-z/x-3y+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
x+2y=3 x+y+z=36 2x+y+z=15 2y=3z x-y=1 x+2y+z x-z=-1 2x+z-y=1
(x+y-z)(x-y+z)=
已知(x+y+z)^2=x^2+y^2+z^2,证明x(y+z)+y(z+x)+z(x+y)=0
已知x、y、z满足方程组:x+y-z=6;y+z-x=2;z+x-y=0 求x、y、z的值