x+y=2z=4,x-2y+z=-2,x+2y+3z=0

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=3x+2z=4y+z 求x:y:z x+y/2=z+x/3=y+z/4 x+y+z=27 若x,y,z成等差数列,则(z-x)^2-4(x-y)(y-z)= (x*x+2)(y*y+4)(z*z+8)=64xyz,求x,y,z (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 2x+z=10 x+y-z=4 3x-y-z=0 已知整数x、y、z,满足x≤y＜z,且｜x+y｜+｜y+z｜+｜z+x｜=4,｜x-y｜+｜y-z｜+｜z-x｜=2,求x^2+y^2+z^2的值. 整数x、y、z满足x≤y＜z|x+y|+|y+z|+|z+x|=4|x-y|+|y-z|+|z-x|=2求x²+y²+z² 已知整数x,y,z满足x≤y＜z,且|x+y|+|y+z|+|z+x|=4 |x-y|+|y-z|+|z-x|=2 那么x²+y²+z²的值 (x+y-z)^2-(x-y+z)^2=? (x-y-z)*( )=x^2-(y+z)^2 填空 x=y/z=z/3,x+y+z ＝12,求2x+3y+4z是多少, 若x,y,z成等差数列,则(z-x)^2-4(x-y)(y-x)=