3x+2y=14,x+y+z=10,2x+3y-z=1

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 (4x-2y-z)-{5x-[8y-2z-(x-2y)]-x-(3y-10z)}=? [3x+2y+z=14,x+y+z=10,2x+3y-z=1] 3x+2y+z=14 { x+y+z=10 2x+3y-z=1 3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5 三元一次方程组,1.x-2y+z=9 2x+y+3z=10 3x+2y1.x-2y+z=92x+y+3z=103x+2y-z2.3x-5y+2z=14x+5y-z=5X-5y+3z=-23.7x+2y-z=76x-y+z=74x+3y+z=94.4x+3y-2z=-75x-6y+z=12x+z=-25.2x+3y-4z=-23x-2y+5z=262x+5z=31 (4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuaihuajian X=3z X+y+z=12 X+2y-2z=10 2x+z=10 x+y-z=4 3x-y-z=0 2x+y-z=10 ,x+3y+2x=14 ,3x-y-2z=7 x+y+z=6 x+2y+3z=14 y+1=z X/10=Y/8=Z/9,则X+Y+Z/2Y-3Z 3X+2y+z=20 x+y+z=10 Y=3X {2x+y=10 {x-y+z=4 {3x-y-z=0 证明 ：x/(y+z)+y/(z+x)+z/(x+y)>=3/2其中 x,y,z>0 (4x-2y-z)-{5x-[8y-2z-(x+y)]-2(3y-10z)}=?好难啊 x+y+z=14 7z=x+y+2 x+z=y