Đáp án:
\[\left[ \begin{array}{l}
x = 0\\
x = \frac{{28 \pm \sqrt {409} }}{5}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {\frac{{x + 3}}{{x - 5}}} \right)^2} + 6.{\left( {\frac{{x - 3}}{{x + 5}}} \right)^2} = 7.\frac{{{x^2} - 9}}{{{x^2} - 25}}\\
\Leftrightarrow {\left( {\frac{{x + 3}}{{x - 5}}} \right)^2} - 7.\frac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} + 6.{\left( {\frac{{x - 3}}{{x + 5}}} \right)^2} = 0\\
\Leftrightarrow {\left( {\frac{{x + 3}}{{x - 5}}} \right)^2} - 7.\frac{{x + 3}}{{x - 5}}.\frac{{x - 3}}{{x + 5}} + 6{\left( {\frac{{x - 3}}{{x + 5}}} \right)^2} = 0\\
\Leftrightarrow \left( {\frac{{x + 3}}{{x - 5}} - \frac{{x - 3}}{{x + 5}}} \right)\left( {\frac{{x + 3}}{{x - 5}} - 6.\frac{{x - 3}}{{x + 5}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\frac{{x + 3}}{{x - 5}} = \frac{{x - 3}}{{x + 5}}\\
\frac{{x + 3}}{{x - 5}} = 6.\frac{{x - 3}}{{x + 5}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + 8x + 15 = {x^2} - 8x + 15\\
{x^2} + 8x + 15 = 6.\left( {{x^2} - 8x + 15} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
5{x^2} - 56x + 75 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \frac{{28 \pm \sqrt {409} }}{5}
\end{array} \right.
\end{array}\)
