设z=xφ(x+y)+yψ(x+y),其中φ,ψ具有二阶连续导数,验证x^2*x的二阶偏导-2*混合偏导(先x后y)+y的二阶偏导=0

设z=z(x,y)由方程φ(x/z,y/z) 确定,证明x*∂z/∂x+y*∂z/∂y=z 设z=x/(x+y),求二阶偏导 设Z=X+Y,其中X,Y满足X+2Y>=0,X-Y 设z=(x+y)/(x-y),求dz (x+y-z)(x-y+z)= 设z=f(x,y) 设z=x^y,求dz 设z=ln(x^z×y^x),求dz 设z={x-y x≥2y,x/4+y/2 x 设x,y,z为正数,且xyz(x+y+z)=4,则(x+y)(y+z)的最小值 设|x|-|y|+|z|=|x+y+z|,且|x+y|=4,|y+z|=5,|y|=3,求(x-y-z)^2的值 设方程z^y=y^x 确定函数z=z(x,y) ,求∂z/∂x X+Y+Z=? x/y=(x+z)/(y+z)y/z=(x+y)/(x+z) 已知 x/(y+z)+y/(z+x)+z/(x+y)=1求 (x*x)/(y+z)+(y*y)/(x+z)+(z*z)/(x+y)=? z+=x>y?++x: 设f(x+y,x-y)=xy,z=(xy,x/y),则dz= (y+z)/x=(z+x)/y=(x+y)/z求x+y-z/x+y+z的值