(1/x^2 + 1/y^2 + 1/z^2) 在 x^2+y^2+z^2=1 内 求 三重积分小弟在此膜拜

1、y-x/x²-y²2、(x-y)(y-z)(z-x)/(z-y)(y-x)(x-z) 已知实数x y z满足x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3,求x+y+z的值 已知实数x y z满足x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3,求x+y+z的值 x+y+z=6 x+2y+3z=14 y+1=z 若|x+3|+|y-2|+|2×z+1|=0求(x×z-y×z)(y-x+z)的值 matlab解对数方程组[x,y,z]=solve('0.29744*x/(3+0.1*x+y+z)=(log10((3+x+y+z)/(3+0.1*x+y+z)))^2','0.79*x/(3+0.1*y+z)=(log10((3+x+y+z)/(3+0.1*x+y+z)*(3+y+z)/(3+0.1*y+z)))^2','1.25*x/(3+0.1*z)=(log10((3+x+y+z)/(3+0.1*x+y+z)*(3+y+z)/(3+0.*1y+z)*(3+z)/ {x+y+z=1;x+3y+7z=-1;z+5y+8z=-2 2|X-Y|+√2Y+Z+Z²-Z+1/4=0求X+Y+Z的值 因式分解x^2y^2z^2+x^2z+y^2z+1 如果,根号x-3+| y-2 |+z^2=2z-1 求 （x+z)^y x-z=-4 z-2y=-1 x+y-z=-1 怎么解 x+y+z=2,x-z=-2,y+z=1 {X+y+z=8,X一z=-2,y十z=1 分式加减的几道题目求速度1 .(x+y)²/（x-z)(y-z) - (y+z)²/（x-z)(y-z) = 2.2x+3y/x-y+2y/y-x-3y/x-y= 设x,y.z为实数,2x.3y.4z是等比数列,1/x,1/y.1/z是等差数列,则(x/z)+(z/x)是多少? 已知x,y,z满足 2倍(x-y的绝对值)+(2y的平方根)-z+(z×z)=-1/4,求x,y,z的值 已知x,y,z>0,xyz(x+y+z)=1,求证（x+y）(x+z)>=2 X+2Y= -12X+Y+Z= 1-X+Y-Z= -1