x+y=8,x+z=10,y+x=12,那么x+2y+x=?

(x+y-z)(x-y+z)= (4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai f(x,y,z,w)=x*(x+y)*(x+y+z)*(x+y+z+w) X+Y+Z=? 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=8,y+z=12,x+z=10,那么x+2y+z等于多少呢? 如果|x+y+z-6|+|2x+3y-z-12|+|2x-y-z|=0求x,y, y=-2X,Z=3y,那么X-Y+(-Z)等于 x,y,z为实数 且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+z-2y)^2+(x+y x,y,z为实数且(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y- 2X+Y+Z=10,X+2Y+Z=-6,X+Y+2Z=8 5X+4Y-3Y=13,2X-3Y+7Z=19,3X-Y=2X x,y,z分别代表三个不同的数,且X+X=x=y+y,y+y+y+y=z+z+z,x+y+y+z=60 ,那么Y+X+Z