求Z=X (X^2 Y^2)的二阶偏导

√x+√(y-1)+√(z-2)=1/2(x+y+z)变形后得[x-2√x+1]+[(y-1)-2√(y-1)+1]+[(z-2)-2√(z-2)+1=0即（√x-1）^2+[√(y-1)+1]^2+

令u=x-y,v=y/xaz/ax＝az/au×au/ax＋az/av×av/ax＝fu-y/x^2×fva^2z/axay＝a(az/ax)/ay＝a(fu-y/x^2×fv)/ay＝a(fu)/a

设x/3=y/4=z/5=k,则x=3ky=4kz=5k带入x+y+z/3x-2y+z,最后约去k就可以了

x+y-z=6y+z-x=2z+x-y=0三式相加得x+y+z=8-得2z=2z=1-得2x=6x=3-得2y=8y=4x=3y=4z=1

x:y:z=4:5:7令x=4k,y=5k,z=7k(2x+3y+z)/5z=(2*4k+3*5k+7k)/5*7k=30k/35k=6/7(x+y)/(y+z)=(4k+5k)/(5k+7k)=9k

设x/3=y/4=z/5=m则x=3m,y=4m,z=5m则x+y+z/3x-2y+z=（3m+4m+5m）/（9m-8m+5m）=12/6=2

因为：X+Y+Z=0得：Z+Y=-X------(1)X+Y=-Z------------(2)Z+Y=-X------------(3)X^3+X^2Z-XYZ+Y^2Z+Y^3=X^3+XZ(X+

因为x:y:z=2:3:4所以设x=2k,y=3kz=4k因为x-y+z=36所以：2k-3k+4k=363k=36k=12所以x+y+z=2k+3k+4k=9k=9×12=108

因为2x=3y=4z可得,x=2z,y=4/3z代入2x-y+z=11/3zx+y+z=13/3z2x-y+z分之x+y+z的值等于13/11

设：x/4=y/5=z/6=k则有：x=4k,y=5k,z=6k(x+y+z)/(3x-2y+z)=(4k+5k+6k)/(12k-10k+6k)=15k/8k=15/8

由柯西不等式(a^2+b^2+c^2)(x^2+y^2+z^2)>=(ax+by+cz)^2,得((1/√2)^2+(1/√3)^2+1)(2x^2+3y^2+z^2)>=(x+y+z)^22x^2+

x:y:z=2:3:4令x=2k,y=3k,z=4k(x+y+z)/(x-2y+3z)=(2k+3k+4k)/(2k-2*3k+3*4k)=9k/(8k)=9/8

因为x:y:z=3：4：5所以设x=3k,y=4k,z=5k(k≠0)(1)z/(x+y)=5k/(3k+4k)=5k/7k=5/7(2)x+y+z=63k+4k+5k=612k=6k=1/2x=3k

8X+Y-Z=m(X+2Y-Z)+n(2X-Y+Z)=(m+2n)X+(2m-n)Y+(-m+n)Zm+2n=8;2m-n=1;-m+n=1解得,m=2,n=3所以8X+Y-Z=2×8+3×18＝70

设x=2t,则y=3t,z=4t3x+2y-z/x+y+z=6t+6t-4t/2t+3t+4t=8t/9t=8/9再问：x：y：z=2：3：4这个可以用两内项之级等于两外项之级吗？？？？再答：可以的。

(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)&

答：x=2,x+y+z=-2.82+y+z=-2.8y+z=-4.8x²(-y-z)-3.2x(z+y)=-x(y+z)(x+3.2)=-2×(-4.8)×(2+3.2)=9.6×5.4=5

2X^2+2Y^2+Z^2+8XZ+8=0上式关于x求偏导：4x+2z*z'(关于x的偏导)+8z+8xz‘（关于x的偏导）=0可得出z’（关于x的偏导）,二阶导类似这样进行.

括号是什么意思?只有一半

x*x+y*y+2z*z-2x+4y+4z+7=0(x*x-2x+1)+(y*y+4y+4)+2(z*z+2z+1)=0(x-1)^2+(y+2)^2+2(z+1)^2=0x=1,y=-2,z=-1x