已知z=x^3 y^3-3xy,求极值

x²-6xy+10y²+4y+|z²-3z+2|+4=0(x²-6xy+9y²)+(y²+4y+4)+|z²-3z+2|=0(x-

x+y=5x=5-yz^2=xy+y-9z^2=(5-y)y+y-9z^2=-y^2+6y-9z^2=-(y-3)^2z^2+(y-3)^2=0所以,z=0,y-3=0z=0,y=3x=5-y=5-3

xy+yz+xz={（x²+y²+z²+2xy+2xz+2yz）-（x²+y²+z²）}\2={（x+y+z）²-（x²

答：x+y+z=3y=2zy≠0,则z≠0所以：y=2z/3x+2z/3+z=2zx=z/3令z=3k,y=2k,x=k(xy+yz+zx)/(x²+y²+z²)=(2k

x-y=5x=5+yz^2=-xy-y-9=-(5+y)y-y-9=-y^2-6y-9=-(y+3)^2所以,z=0,y+3=0z=0,y=-3x=5+y=5-3=2x-2y+3z=2-2*(-3)+

已知x^3+y^3-z^3=96,xyz=4,x^2+y^2+z^2-xy+xz+yz=12,则x+y-z等于[x+y-z]^2=x^2+y^2+z^2-2xy-2xz-2yzx^3+y^3=（x+y

4x-3y=3z.(1)x-3y=z.(2)(1)-(2)得3x=2zx=(2/3)z代入(2)得(2/3)z-3y=zy=-(1/9)z则xy+2yz/x²+y²+z²

∵x+y=5，z2=xy+y-9，∴x=5-y，代入z2=xy+y-9得：z2=（5-y）y+y-9，z2+（y-3）2=0，z=0，y-3=0，∴y=3，x=5-3=2，x+2y+3z=2+2×3+

y=-12;一共是三个方程,因为xy/(x+y)=3推出（x+y）/(xy)=1/3-------方程1；同理：（y+z）/(yz)=1/2-------方程2；（x+z）/(xz)=1-------

y=4/3XZ=2X带进去算就可以了

x=5-yz2=(5-y)y+y-9=6y-y2-9=-(9-6y+y2)=-(y-3)2由题意,只有当该项为0时等式成立得y=3那么z=0x=2即原式=2+6+0=8

1.z=3y/2把:z=3y/2代入x+y+z=3y得:x+y+3y/2=3y整理后得:x=y/2所以:x/(x+y+z)=(y/2)/(y/2+y+3y/2)=1/62.因为1/x-1/y=3,则1

令X=3k,由于x:y:z=3:4:6则:y=4k,z=6k将x=3k,y=4k,z=6k代入（xy+yz+xz）/（x^+y^+z^）则有：（xy+yz+xz）/（x^+y^+z^）=（12k^+2

xy/(x+y)=1=>xy=x+y=>1/x+1/y=1--式一yz/(y+z)=2=>yz=2y+2z=>1/y+1/z=1/2--式二xz/(x+z)=3=>xz=3x+3z=>1/x+1/z=

（x+y+z）^2=x^2+y^2+z^2+2(xy+yz+zx)=x^2+y^2+z^2+2=(x^2+y^2)/2+(y^2+z^2)/2+(x^2+z^2)/2+2≥2[√(x^2*y^2)]/

令2/x=3/y=7/z=k∴x=2/ky=3/kz=7/k∴(xy+xz+yz)/(x^2+y^2+z^2)=(2/k*3/k+2/k*7/k+3/k*7/k)/(4/k²+9/k

解方程组：｛2x-3y-z=0.(1)｛x+3y-14z=0.(2)(1)+(2)得：3x-15z=0即：x=5z,代入（1）式得y=3z所以：(4x²-5xy+z²)/(xy+y

解题思路：本题的关键是将三个方程两边取倒数，化简后分别将方程等号左边和右边相加，得到1/x+1/y+1/z的值，最后将要求的分式化简，把1/x+1/y+1/z的值带入即可。解题过程：

3x-y=-2zx+2y=-3z那么：x=-z,y=-z（3x^－xy+2y^)/(2x^+4xy+y^)=(3z^2-z^2+2z^2)/(2z^2+4z^2+z^2)=4z^2/7z^2=4/7