由X^2+Y^2+Z^2-2X+4Y+1/2=0
由X^2+Y^2+Z^2-2X+4Y+1/2=0
3道高数题,1,函数F(x,y,z)=(e^x) * y * (z^2) ,其中z=z(x,y)是由x+y+z+xyz=
请问怎么由2x-3y+z=0 ,3x-2y-6z=0 得到x=3z,y=4z
设函数z=z(x,y)由方程x^2+y^3-xyz^1=0确定,求z/x,z/y
x^2+y^2+z^2+4x+4y+4z+1=0,求x+y+z
x^2+y^2+z^2+4x+4y+4z+1=0求x+y+z
x^2+y^2+z^2+4x+4y+4z+1=0 求x+y+z
1.已知x,y,z满足2│x-y│+(根号2y-z)+z平方-z+(1/4)=0,求x,y,z值.
由方程组{x-2y+3z=0 可得x:y:z是多少 {2x-3y+4z=0
已知 x,y,z都是正实数,且 x+y+z=xyz 证明 (y+x)/z+(y+z)/x+(z+x)/y≥2(1/x+1
(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z
已知3x-2y-5z=0,2x-5y+4z=0,且x,y,z均不为0,求3x*x+2y*y+5z*z/5x*x+y*y-