x+y+z=7,2x+3y+4y=20求x,y,z的正整数解
求方程组x+y+z=7,2x+3y+4z=20的正整数解
求满足下列条件的所有正整数解3x+2y-z=4,2x-y+2z=6,x+y+z<7
已知3x+2y=4+z,2x+2z=6+y,问是否存在x、y、z的正整数值,使得x+y+z
已知x+4y+3z=3x-2y-5z=0,求x+2y-z/2x-3y+7z的值
已知4x-3y-6y=0,x+2y-7z=0.求x-y+z\x+y+z的值
x+2y+3z=9,求正整数x、y、z
2x+3y+4z=20;x+y+z=7 的解
求方程组 {x+y+z=12,x+2y-3z=2}的正整数解
求方程组x+y-z=1,2x+3y-z=13的正整数解
已知x/2=y/4=z/7,求①x:y:z.②若2x+y+3z=58,分别求x,y,z的值.③求x+y/x+z的值.
x y z x+y--- = --- = ---- ----y+Z z+x x+y ,求 z 的值 .求 x+y----
x+2y+3z=20,x+3y+5z=31,求x+y+z的值