{x+y+z=6,x-z=2,2x-y+z=5 解方程组

用行列式的性质证明：y+z z+x x+y x y z x+y y+z z+x =2 z x y z+x x+y y+z y z x 这个怎么证? (x-2y+z)(x+y-2z)分之（y-x)(z-x) + (x+y-2z)(y+z-2x)分之(z-y)(x-y) + (y+z-2z)(x-2y+z)分之（x-z)(y-z)=?第三部分那个是 （y+z-2x)(x-2y+z)分之（x-z)(y-z) 试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) 已知(x+y)(x+z)=x,(y+z)(y+x)=2y,(z+x)(z+y)=3z,求x,y,z 已知：x^2/(z+y)+y^2/(x+z)+z^2/(x+y)=0,求x/(z+y)+y/(x+z)+z/(x+y)的值. x^2/(z+y)+y^2/(x+z)+z^2/(x+y)=0,求x/(z+y)+y/(x+z)+z/(x+y)的值 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 已知X.Y.Z满足方程组,X+y-Z=6y+z-x=2z+x-y=0求X.Y.Z的值 方程组：{x+y-z=6,x+z=16,x-y+2z=17 x+y+z=6 x+2y+3z=14 y+1=z 如果|x+y+z-6|+|2x+3y-z-12|+|2x-y-z|=0求x,y, (x+y-z)^2-(x-y+z)^2=? (x-y-z)*( )=x^2-(y+z)^2 填空 分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y) 证明 ：x/(y+z)+y/(z+x)+z/(x+y)>=3/2其中 x,y,z>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 (x+y-z)(x-y+z)= x+2y=3x+2z=4y+z 求x:y:z