Question

Cho các số: x=by+cz

y=ax+cz

z= ax+by và x+y+z khác 0

Tính Q=\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\)

Answer 1

\(x+y=by+cz+ax+cz=ax+by+2cz=z+2cz\)

\(\Rightarrow2cz=x+y-z\Rightarrow c=\frac{x+y-z}{2z}\Rightarrow c+1=\frac{x+y-z}{2z}+1=\frac{x+y+z}{2z}\)

\(\Rightarrow\frac{1}{1+c}=\frac{2z}{x+y+z}\)

Tương tự ta có: \(\frac{1}{1+a}=\frac{2x}{x+y+z}\) ; \(\frac{1}{1+b}=\frac{2y}{x+y+z}\)

\(\Rightarrow Q=\frac{2x}{x+y+z}+\frac{2y}{x+y+z}+\frac{2z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)