Tính 3a + 3b. Biết \(?\left\{ \begin{array}{l}\left( {1 + \sqrt 3 } \right)x + \left( {\sqrt 2 - 1} \right)y = 1\\\left( {1 + \sqrt 2 } \right)x + \left( {1 - \sqrt 3 } \right)y = 1\end{array} \right.\) có nghiệm là (a;b).

 \(\begin{array}{l}\,\,\left\{ \begin{array}{l}\left( {1 + \sqrt 3 } \right)x + \left( {\sqrt 2 - 1} \right)y = 1\\\left( {1 + \sqrt 2 } \right)x + \left( {1 - \sqrt 3 } \right)y = 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left( {1 + \sqrt 3 } \right)\left( {1 + \sqrt 2 } \right)x + y = \sqrt 2 + 1\\\left( {1 + \sqrt 2 } \right)\left( {1 + \sqrt 3 } \right)x - 2y = 1 + \sqrt 3 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}3y = \sqrt 2 - \sqrt 3 \\\left( {1 + \sqrt 3 } \right)x + \left( {\sqrt 2 - 1} \right)y = 1\end{array} \right. \\ \Leftrightarrow \left\{ \begin{array}{l}y = \dfrac{{\sqrt 2 - \sqrt 3 }}{3}\\\left( {1 + \sqrt 3 } \right)x + \dfrac{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 - \sqrt 3 } \right)}}{3} = 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}y = \dfrac{{\sqrt 2 - \sqrt 3 }}{3}\\\left( {1 + \sqrt 3 } \right)x + \dfrac{{2 - \sqrt 6 - \sqrt 2 + \sqrt 3 }}{3} = 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}y = \dfrac{{\sqrt 2 - \sqrt 3 }}{3}\\\left( {1 + \sqrt 3 } \right)x = \dfrac{{1 + \sqrt 6 + \sqrt 2 - \sqrt 3 }}{3}\end{array} \right.\\ \\\Leftrightarrow \left\{ \begin{array}{l}y = \dfrac{{\sqrt 2 - \sqrt 3 }}{3}\\x = \dfrac{{1 + \sqrt 6 + \sqrt 2 - \sqrt 3 }}{{3\left( {1 + \sqrt 3 } \right)}}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}y = \dfrac{{\sqrt 2 - \sqrt 3 }}{3}\\x = \dfrac{{2\sqrt 3 + 2\sqrt 2 - 4}}{6}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}y = \dfrac{{\sqrt 2 - \sqrt 3 }}{3}\\x = \dfrac{{\sqrt 3 + \sqrt 2 - 2}}{3}\end{array} \right.\end{array}\)

⇒ 3a + 3b = \(2\sqrt2-2\)