z=x^2+y^2,z^2=x^2+y^2,z=根号下x^2+y^2,他们的图形分别是什么?

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^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)(x+z)=x,(y+z)(y+x)=2y,(z+x)(z+y)=3z,求x,y,z (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 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=3y=2z,则x/x+y+z=? x+y+z=14 7z=x+y+2 x+z=y x/2=y/3=z/5 x+3y-z/x-3y+z 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>=y>=z>0,求证：x^2*y/z+y^2*z/x+z^2*x/y>=x^2+y^2+z^2