先去分母化成整数方程:x+y+z=3200 20x+15y+12z=48000 12x+15y+20z=52800 救命
20(x+y)=12(y+z)=15(z+x),且xyz不等于零,求x:y:z=
(x+y-z)(x-y+z)=
已知20(x+2y)=15(y+3Z)=12(z+5x),则x:y
解方程组{x(x+y+z)=6,y(x+y+z)=12,z(x+y+z)=18
x+y=10 y+z=15 z+x=20 求x,y,z等于多少?
解方程{3x -y+z=4 2x+3y-z=12 x+y+z=6
3x-y+z=3 2x+y-3z=11 x+y+z=12 解方程
解方程x+y+2z=7 2x+3y-z=12 3x+2y+z=13 时宜先消去未知数
X+Y+Z=?
x+y+z=14,x'+y'+z'=15,(x-x')+(y+y')+z*z'=16,已知xyz为自然数,求x,y,z的
xy/x+y=12,xz/x+z=15,yz/y+z=20,求x,y,z,的值
设X,Y,Z都是整数,满足条件(X-Y)(Y-Z)(Z-X)=X+Y+Z,试证明X+Y+Z能被27整除