x+2y+z=1 x+y+2z=2 2x+y+z=5
2个回答
3式相加,得
x+y+z=2
所以,容易得,
x=3,y=-1,z=0
相关问题
解下列方程组:(1){2x+y-z=2,x+2y-z=5,x-y+z=-5;(2){2x+y+z=15,x+2y-z=5
计算:(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(
1.x-y=5x+z=-7y-z=-82.3x+2y+5z=2x-2y-z=64x+2y-7z=303.2x+y=33x
x^2(y+z)^2-2xy(x-z)(y+z)+y^2(x-z)^2 =[x(y+z)-y(x-z)]^2 =(xz+
x+y+2z=0 2X-y+z=1 x+2y-3z=5
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/
1,{2x+y+z=-26,x+2y+z=-30,x+y+2z=-28 2,{5x+2y+z=12,4x+3y+z=13
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