x+y/2=z+x/3=y+z/4 x+y+z=27
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x+y/2=z+x/3=y+z/4 x+y+z=27
x+y/2=z+x/3=y+z/4 ①
x+y+z=27 ②
①×12得:12x+6y=12z+4x=12y+3z
故:
12x+6y=12z+4x ③
12z+4x=12y+3z ④
由③得 x=(6z-3y)/4
把 x=(6z-3y)/4代入②和④中:
(6z-3y)/4+y+z=27
12z+4×(6z-3y)/4=12y+3z
解上述二元一次方程组得:y=z=108/11
将y=z=108/11代入②中得:x=81/11
∴x=81/11
y=108/11
z=108/11
x+y+z=27 ②
①×12得:12x+6y=12z+4x=12y+3z
故:
12x+6y=12z+4x ③
12z+4x=12y+3z ④
由③得 x=(6z-3y)/4
把 x=(6z-3y)/4代入②和④中:
(6z-3y)/4+y+z=27
12z+4×(6z-3y)/4=12y+3z
解上述二元一次方程组得:y=z=108/11
将y=z=108/11代入②中得:x=81/11
∴x=81/11
y=108/11
z=108/11
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
(x+y-z)(x-y+z)=
x=y/z=z/3,x+y+z =12,求2x+3y+4z是多少,
求解三元一次方程组:{x+y/2=y+z/3=z+x/4 x+y+z=27
x/2=y/3=z/5 x+3y-z/x-3y+z
X+Y+Z=?
已知x,y,z 大于0,x+y+z=2,求证 xz/y(y+z)+zy/x(x+y)+yx/z(z+x)大于等于2/3
1.x+y=16,y+z=12,z+x=102.3x-y+z=4,2x+3y-z=12,x+y+z=63.x+y+z=6
(x+y-z)^2-(x-y+z)^2=?
(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai
x+y+z=2 4x+2y+z=4 2x+3y+z=1
f(x,y,z,w)=x*(x+y)*(x+y+z)*(x+y+z+w)