Question

giải 

\(\hept{\begin{cases}x^2+2xy=x\\y^2+2xz=y\\z^2+2xy=z\end{cases}}\)

Answer 1

Ta có: \(\hept{\begin{cases}x^2+2xy=x\\y^2+2xy=y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\left(x+2y\right)=x\\y\left(y+2x\right)=y\end{cases}\Leftrightarrow\hept{\begin{cases}x+2y=1\\y+2x=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x+2y=1\\4x+2y=2\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}-3x=-1\\x+2y=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{3}\\\frac{1}{3}+2y=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=\frac{1}{3}\end{cases}}}\)