x+2y-z=1,2x+y-2z=2则x+y-z=
4个回答
两式相加
x+2y-z+2x+y-2z=1+2
3x+3y-3z=3
x+y-z=1
在我回答的右上角点击【评价】,然后就可以选择【满意,问题已经完美解决】了.
相关问题
x,y,z为正实数,则( z^2-x^2)/(x+y)+(x^2-y^2)/(y+z)+(y^2-z^2)/(z+x)的
计算:(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(
x^2(y+z)^2-2xy(x-z)(y+z)+y^2(x-z)^2 =[x(y+z)-y(x-z)]^2 =(xz+
已知x,y,z满足x/(y+z)+y/(z+x)+z/(x+y)=1,求代数式x2/(y+z)+y2/(x+z)+z2/
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(xy-2z)(y+z-2x)+(x-
-(x-y+z)-2(x-y+z)-3(x-y+z),其中x-=1,y=1/2,z=-2
若x+y+z=2,x^2+y^2+z^2=2,1/x+1/y+1/z=1/3则 x^3+y^3+z^3=
1.x+y+z=21,x-y=1,2x+z-y=13.2.3x+2y+z=13,x+y+2z=7 ,2z+3y-z=12
1.因式分解x^2 y - y^2 z + z^2 x-x^2 z +y^2 x + z^2 y -2xyz
x,y,z为实数 且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+z-2y)^2+(x+y