方程只要答案:1,x+y+z=2 2y-3z=1 2x-y+3z=-12,已知(x-y+2z)/3=(y-z+2x)/7
3个回答
1.x= -1 y=2 z=1
2.x= 3 y=2 z=1
相关问题
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
2x-y=7 x+y+z=25x+3y+z=2 x-2y+z=-1 3x-4z=4 x+2y+3z=-1x+y-z=6x
{3x+2y+z=13 x+y+2z=7 2x+3y-z=12,求x,y,z
证明:(y+z-2x)3+(z+x-2y)3+(x+y-2z)3=3(y+z-2x)(z+x-2y)(x+y-2z).
{3x+2y+z=13 {x+y+z=7 {2x+3y-z=12
x-5y-3z=0 y+z-3x=3 x+y+3z=14 y+4z=3 z+x-3y=5 2x+y+z=7 2x-z=1
已知x+y+z=1,x2+y2+z2=2,x3+y3+z3=3,求xy(x+y)+yz(y+z)+zx(z+x)的值
1、已知x/y=7/2,求x/x+y 2、已知x/2=y/3=z/4,求x+y+z/2x+3y-z!
已知x+y+z=1 x2+y2+z2=2 x3+y3+z3=3 求x4+y4+z4=?
已知方程x:y:z=2:1:3,则[2x−y+3z/x+2y]=______.