Đáp án:
\[\left[ \begin{array}{l}
x = 1\\
x = \frac{{27 + 10\sqrt 2 }}{{23}}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left( {5 - \sqrt 2 } \right){x^2} - 10x + 5 + \sqrt 2 = 0\\
\Leftrightarrow \left[ {\left( {5 - \sqrt 2 } \right){x^2} - \left( {5 - \sqrt 2 } \right)x} \right] - \left[ {\left( {5 + \sqrt 2 } \right)x - \left( {5 + \sqrt 2 } \right)} \right] = 0\\
\Leftrightarrow \left( {5 - \sqrt 2 } \right)x\left( {x - 1} \right) - \left( {5 + \sqrt 2 } \right)\left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left[ {\left( {5 - \sqrt 2 } \right)x - \left( {5 + \sqrt 2 } \right)} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{{5 + \sqrt 2 }}{{5 - \sqrt 2 }}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{{{{\left( {5 + \sqrt 2 } \right)}^2}}}{{\left( {5 - \sqrt 2 } \right)\left( {5 + \sqrt 2 } \right)}} = \frac{{27 + 10\sqrt 2 }}{{23}}
\end{array} \right.
\end{array}\)
