求解此方程组 x(x+y+z)=6.y(x+y+z)=12.z(x+y+z)=18

解方程组{x(x+y+z)=6,y(x+y+z)=12,z(x+y+z)=18 (x+y-z)(x-y+z)= 求解三元一次方程组：｛x+y/2=y+z/3=z+x/4 x+y+z=27 X+Y+Z=? 已知x、y、z满足方程组：x+y-z=6；y+z-x=2；z+x-y=0 求x、y、z的值 f(x,y,z,w)=x*(x+y)*(x+y+z)*(x+y+z+w) 解方程组x-4y+z=-3,2x+y-z=18,x-y-z=7. 解下列方程组x+y+z=18 x+y-z=0 x-2y+z=0 x y z x+y--- = --- = ---- ----y+Z z+x x+y ,求 z 的值 .求 x+y---- 已知x+y-z/z=x-y+z/y=-x+y+z/x,且xyz不等于0,求分式[(x+y)(x+z)(y+z)]/xyz 若z分之x+y+z=y分之x-y+z=x分之-x+y+z,求xyz分之(x+y)(y+z)(z+x) y+z÷x=Z+X÷y=X+Y÷z,X+Y+Z不等0求X+Y-Z÷X+Y+z值