z=x y㏑(2x² 3y²),求偏导

实数x,y,z,满足那么x+y=6,z^2=xy-9,∴xy=z^+9,(x-y)^=(x+y)^-4xy=-4z^>=0,∴z=0,(x+y)^z=6^0=1.

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

x-3=y-2x-y=1y-2=z-1y-z=1x-3=z-1z-x=-2x^2+y^2+z^2-xy-yz-xz=x(x-y)+y(y-z)+z(z-x)=x+y-2zx-3=z-1y-2=z-12

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

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

题目出错了吧应该是x^2-4x+y^2+6y+√(z-3)+13=0即（x-2)^2+(y+3)^2+√(z-3)=0(xy)^z=(2*(-3))^3=-216啊你也去想想吧

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)+

(xy+yz+xz)²=x²y²+x²z²+y²z²+2xyz²+2x²yz+2xy²z=1=x&#

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-------

令x/3=y/2=z/5=k则x=3ky=2kz=5k∴(xy+yz+zx)/(x²+y²+z²)=(6+10+15)k²/(9+4+25)l²=31

3[-(x+y)+2xy²-z]-2[(x+y)-xy²+z]-5[-3(x+y)-z]=3(-x-y+2xy²-z)-2(x+y-xy²+z)-5(-3x-3

(x+y+z)²+(x-y-z)(x-y+z)-2·z(x+y)=(x+y)²+2z(x+y)+z²+(x-y)²-z²-2z(x+y)=(x+y)&

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

1/Y+1/X=1（1）1/Z+1/Y=2（2）1/X+1/Z=3（3）（1）+（2）+（3）：1/X+1/Y+1/Z=3（4）（4）-（1）：1/Z=2Z=1/2（4）-（2）：1/X=1X=1题目

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：

｜x-3｜+｜y+z｜+｜2z+1｜=0则｜x-3｜=0x=3｜y+z｜=0y=-z=1/2｜2z+1｜=0z=-1/2xy-yz=3x1/2-1/2x(-1/2)=7/4

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