x=by+cz,y=ax+cz,z=ax+by
=>x+y+z=2(ax+by+cz) (1)
Thay z=ax+by vào (1) ta có :
x+y+z=2(z+cz)=2z(c+1)
\(=>\frac{1}{c+1}=\frac{2z}{x+y+z}\)
Tương tự ta có : \(\frac{1}{a+1}=\frac{2x}{x+y+z},\frac{1}{b+1}=\frac{2y}{x+y+z}\)
=>Q=\(\frac{2\left(x+y+z\right)}{x+y+z}=2\)