x^2+y^2+z^2=2z,φ(x,y)=0且具有连续偏导数,求dz/dx

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 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(x-2z+y)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z) 计算:(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z) 计算:x^2/(x-y)(x-z)+y^2/(y-x)(y-z)+z^2/(z-x)(z-y) (x+y-z)^2-(x-y+z)^2=? (x-y-z)*( )=x^2-(y+z)^2 填空 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(xy-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)速速回答 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 化简(x+Y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z) 计算题(x+y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z). 1、y-x/x²-y²2、(x-y)(y-z)(z-x)/(z-y)(y-x)(x-z) x+2y=3x+2z=4y+z 求x:y:z