Đáp án:
$\begin{array}{l}
1)\left( {\frac{{x + 3}}{{x - 3}} - \frac{{x - 3}}{{x + 3}}} \right).\left( {\frac{1}{3} + \frac{1}{x}} \right)\\
= \frac{{{{\left( {x + 3} \right)}^2} - {{\left( {x - 3} \right)}^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\frac{{x + 3}}{{3x}}\\
= \frac{{{x^2} + 6x + 9 - {x^2} + 6x - 9}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}.\frac{{x + 3}}{{3x}}\\
= \frac{{12x}}{{x - 3}}.\frac{1}{{3x}}\\
= \frac{4}{{x - 3}}\\
2){x^2} - 6x + 9 = 2.\left( {x - 3} \right)\\
\Leftrightarrow {\left( {x - 3} \right)^2} = 2\left( {x - 3} \right)\\
\Leftrightarrow \left( {x - 3} \right)\left( {x - 3 - 2} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 5
\end{array} \right.
\end{array}$
