2x+y+2z=4,3x-2y+z=11,x+y-2z=-2

x+2y=3x+2z=4y+z 求x:y:z x+y/2=z+x/3=y+z/4 x+y+z=27 试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) 已知(x+y)(x+z)=x,(y+z)(y+x)=2y,(z+x)(z+y)=3z,求x,y,z 2x+z=10 x+y-z=4 3x-y-z=0 x=y/z=z/3,x+y+z ＝12,求2x+3y+4z是多少, 2x+y+2z=4,3x-2y+z=11,x+y-2z=-2 2x+y+2z=4 3x-2y+z=11 x+y+2z=-2 x+2y+4z=17 2x+y+z=7 3x+y+2z=11 2x+3y+z=0 3x+y-4z=11 x+y+z=-3 2x+3y+z=0 3x+y-4z=11 x+y+z=-3 1.{x:y:z=3:4:5 2x+3y-5z=-142.{x-y-z=-1 3x+5y+7z=11 4x-y+2z=-13.{x+y=1 x+z=0 2x+y+2z=44.{x-y+z=-3 2x-y-z=-2 -3x+y+z=-15.{x-y+z=-3 3x+y-4z=6 2x+y-z=36.{x-2y+5z=16 2x+y-5z=7 2x-y+z=38 (4x-2y-z)-{5x-[8y-2z-(x-2y)]-x-(3y-10z)}=? x-y-z=-1 3x+5y+7z=11 4x-y+2z=-1 分别求出x=?y=?z=? 三元一次方程 X-y-z=-1 3x+5y+7z=11 4x-y+2z=-1 X=?Y=?Z=? 证明 ：x/(y+z)+y/(z+x)+z/(x+y)>=3/2其中 x,y,z>0 x/2=y/3=z/5 x+3y-z/x-3y+z 若x+y+z=3y=2z,则x/x+y+z=?