Giải thích các bước giải:
\(\begin{array}{l}
1,\\
{\rm{DK:}}\,\,\,\left\{ \begin{array}{l}
u \ne \frac{1}{3}\\
u \ne - \frac{{11}}{3}
\end{array} \right.\\
\frac{2}{{\left( {1 - 3u} \right)\left( {3u + 11} \right)}} = \frac{1}{{9{u^2} - 6u + 1}} - \frac{3}{{{{\left( {3u + 11} \right)}^2}}}\\
\Leftrightarrow \frac{2}{{\left( {1 - 3u} \right)\left( {3u + 11} \right)}} = \frac{1}{{{{\left( {1 - 3u} \right)}^2}}} - \frac{3}{{{{\left( {3u + 11} \right)}^2}}}\\
\Leftrightarrow {\left( {\frac{1}{{1 - 3u}}} \right)^2} - 2.\frac{1}{{1 - 3u}}.\frac{1}{{3u + 11}} - 3.{\left( {\frac{1}{{3u + 11}}} \right)^2} = 0\\
\Leftrightarrow \left[ {{{\left( {\frac{1}{{1 - 3u}}} \right)}^2} + \frac{1}{{1 - 3u}}.\frac{1}{{3u + 11}}} \right] - 3.\left[ {\frac{1}{{1 - 3u}}.\frac{1}{{3u + 11}} + {{\left( {\frac{1}{{3u + 11}}} \right)}^2}} \right] = 0\\
\Leftrightarrow \frac{1}{{1 - 3u}}\left( {\frac{1}{{1 - 3u}} + \frac{1}{{3u + 11}}} \right) - 3.\frac{1}{{3u + 11}}.\left( {\frac{1}{{1 - 3u}} + \frac{1}{{3u + 11}}} \right) = 0\\
\Leftrightarrow \left( {\frac{1}{{1 - 3u}} + \frac{1}{{3u + 11}}} \right)\left( {\frac{1}{{1 - 3u}} - 3.\frac{1}{{3u + 11}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\frac{1}{{1 - 3u}} = - \frac{1}{{3u + 11}}\\
\frac{1}{{1 - 3u}} = \frac{3}{{3u + 11}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
1 - 3u = 3u + 11\\
3\left( {1 - 3u} \right) = 3u + 11
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
6u = - 10\\
12u = - 8
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
u = - \frac{5}{3}\\
u = - \frac{2}{3}
\end{array} \right.\\
2,\\
{\rm{DK:}}\,\,\left\{ \begin{array}{l}
x \ne 1\\
x \ne 3
\end{array} \right.\\
\frac{{x + 5}}{{x - 1}} = \frac{{x + 1}}{{x - 3}} - \frac{8}{{{x^2} - 4x + 3}}\\
\Leftrightarrow \frac{{\left( {x + 5} \right)\left( {x - 3} \right)}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} = \frac{{\left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} - \frac{8}{{\left( {x - 1} \right)\left( {x - 3} \right)}}\\
\Leftrightarrow \left( {x + 5} \right)\left( {x - 3} \right) = \left( {x - 1} \right)\left( {x + 1} \right) - 8\\
\Leftrightarrow {x^2} + 2x - 15 = {x^2} - 9\\
\Leftrightarrow 2x = 6\\
\Leftrightarrow x = 3\,\,\,\left( L \right)\\
\Rightarrow ptvn
\end{array}\)
