2x+4Y+8Z=100 x,y,z 各得多少

(x*x+2)(y*y+4)(z*z+8)=64xyz,求x,y,z 2x+4Y+8Z=100 x,y,z 各得多少 (4x-2y-z)-{5x-[8y-2z-(x-2y)]-x-(3y-10z)}=? (4x-2y-z)-{5x-[8y-2z-(x+y)]-2(3y-10z)}=?好难啊 14x+4y+z=200,x+y+z=100求解x,y,z x+2y=3x+2z=4y+z 求x:y:z x+y/2=z+x/3=y+z/4 x+y+z=27 若x,y,z成等差数列,则(z-x)^2-4(x-y)(y-z)= (x+2)*y=z+360 x*(y-8)=z-48 求x/y/z各是多少?最好有解法(x+2)*y=z+360x*(y-8)=z-48求x/y/z各是多少？ X+Y+Z=20,2X+3Y+8Z=100 那么X.Y.Z分别是多少? X+Y+Z=20,2X+3Y+8Z=100 那么X Y Z分别是多少, x:y:z=3:4:5 2x-3y+2z=8 x-y-2z=? 2x+y+z=4 x+2y+z=8 x +y+2z=24 三元一次方程 2x+y+z=4 x+2y+z=8 x+y+2z=12 x+y+z=-1 4x-2y+3z=5 y-z=8-2x 2x+3y+z=8 4x-y-z=10 x+2y-z=1 若x-y+z=-4,x^2-(y+z)^2=8,则x+y+z=如题 x+y+z=51 4x+8y+5z=300 x+y+2z=67