Đáp án:
\(\left[ \begin{array}{l}
x = 0\\
x = 6\\
x = 1\\
x = 5
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:{x^2} - 6x + 7 \ne 0 \to x \ne \left\{ {3 - \sqrt 2 ;3 + \sqrt 2 } \right\}\\
Đặt:{x^2} - 6x - 2 = t\\
Pt \to t + \dfrac{{14}}{{t + 9}} = 0\left( {t \ne - 9} \right)\\
\to t\left( {t + 9} \right) + 14 = 0\\
\to {t^2} + 9t + 14 = 0\\
\to \left( {t + 2} \right)\left( {t + 7} \right) = 0\\
\to \left[ \begin{array}{l}
t = - 2\\
t = - 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} - 6x - 2 = - 2\\
{x^2} - 6x - 2 = - 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
x\left( {x - 6} \right) = 0\\
{x^2} - 6x + 5 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = 6\\
\left( {x - 1} \right)\left( {x - 5} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = 6\\
x = 1\\
x = 5
\end{array} \right.
\end{array}\)
