Giải \(\left\{ \begin{array}{l}3x + 2y = 2\\\left( {2 - \sqrt 5 } \right)x + \left( {1 - \sqrt 5 } \right)y = 2\end{array} \right.\)

* Hướng dẫn giải

Ta có \(\left\{ \begin{array}{l}3x + 2y = 2\\\left( {2 - \sqrt 5 } \right)x + \left( {1 - \sqrt 5 } \right)y = 2\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}y = \dfrac{{2 - 3x}}{2}\\\left( {2 - \sqrt 5 } \right)x + \left( {1 - \sqrt 5 } \right).\dfrac{{2 - 3x}}{2} = 2\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}y = \dfrac{{2 - 3x}}{2}\\\left( {4 - 2\sqrt 5 } \right)x + \left( {1 - \sqrt 5 } \right)\left( {2 - 3x} \right) = 4\end{array} \right. \)\(\Leftrightarrow \left\{ \begin{array}{l}y = \dfrac{{2 - 3x}}{2}\\\left( {4 - 2\sqrt 5 } \right)x - \left( {3 - 3\sqrt 5 } \right)x + 2 - 2\sqrt 5  = 4\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}y = \dfrac{{2 - 3x}}{2}\\\left( {1 + \sqrt 5 } \right)x = 2 + 2\sqrt 5 \end{array} \right. \)\(\Leftrightarrow \left\{ \begin{array}{l}x = 2\\y =  - 2\end{array} \right.\)

Vậy hệ phương trình có nghiệm duy nhất \(\left( {x;y} \right) = \left( {2; - 2} \right)\).