Bài 1: Ta có : \(\left\{ \matrix{ x - 2\sqrt {2y} = \sqrt 5 \hfill \cr \sqrt {2x} + y = 1 - \sqrt {10} \hfill \cr} \right. \)
\(\Leftrightarrow \left\{ \matrix{ \sqrt {2x} - 4y = \sqrt {10} \hfill \cr \sqrt {2x} + y = 1 - \sqrt {10} \hfill \cr} \right.\)
\( \Leftrightarrow \left\{ \matrix{ 5y = 1 - 2\sqrt {10} \hfill \cr x = 2\sqrt {2y} + \sqrt 5 \hfill \cr} \right. \)
\(\Leftrightarrow \left\{ \matrix{ y = {{1 - 2\sqrt {10} } \over 5} \hfill \cr x = {{2\sqrt 2 - 3\sqrt 5 } \over 5} \hfill \cr} \right.\)
Hệ có nghiệm : \(\left( {{{2\sqrt 2 - 3\sqrt 5 } \over 5};{{1 - 2\sqrt {10} } \over 5}} \right).\)
Bài 2: Thế \(x = 3; y = − 2 \) vào hệ đã cho, ta được :
\(\left\{ \matrix{ 3a - 2b = 3 \hfill \cr 6a + 6b = 6 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ 6a - 4b = 6 \hfill \cr 6a + 6b = 6 \hfill \cr} \right. \)
\(\Leftrightarrow \left\{ \matrix{ 10b = 0 \hfill \cr 3a - 2b = 3 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ b = 0 \hfill \cr a = 1. \hfill \cr} \right.\)
