x=y-1,x-2z=1,y-z=-2 这是一个三元一次方程组,求x,y,z.求

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 已知 x/(y+z)+y/(z+x)+z/(x+y)=1求 (x*x)/(y+z)+(y*y)/(x+z)+(z*z)/(x+y)=? 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 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 如果,根号x-3+| y-2 |+z^2=2z-1 求 （x+z)^y 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= 用行列式的性质证明：y+z z+x x+y x y z x+y y+z z+x =2 z x y z+x x+y y+z y z x 这个怎么证? (x-2y+z)(x+y-2z)分之（y-x)(z-x) + (x+y-2z)(y+z-2x)分之(z-y)(x-y) + (y+z-2z)(x-2y+z)分之（x-z)(y-z)=?第三部分那个是 （y+z-2x)(x-2y+z)分之（x-z)(y-z) 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