Đáp án:
$\begin{array}{l}
a)5x + 3 = 3x + 11\\
\Leftrightarrow 2x = 8\\
\Leftrightarrow x = 4\\
b)\frac{{x + 1}}{4} - 2 = \frac{{x - 5}}{5}\\
\Leftrightarrow \frac{{x + 1 - 8}}{4} = \frac{{x - 5}}{5}\\
\Leftrightarrow 5\left( {x - 7} \right) = 4\left( {x - 5} \right)\\
\Leftrightarrow 5x - 35 = 4x - 20\\
\Leftrightarrow x = 15\\
c)\frac{{x - 3}}{{x + 3}} - \frac{2}{{x - 3}} = \frac{{{x^2} + 19}}{{{x^2} - 9}}\left( {dk:x \ne 3;x \ne - 3} \right)\\
\Leftrightarrow \frac{{{{\left( {x - 3} \right)}^2} - 2\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{{x^2} + 19}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
\Rightarrow {x^2} - 6x + 9 - 2x - 6 = {x^2} + 19\\
\Leftrightarrow - 8x = 16\\
\Leftrightarrow x = - 2\left( {tm} \right)\\
d)\frac{1}{{x - 1}} + \frac{1}{{x - 2}} = \frac{1}{{x + 2}} + \frac{1}{{x + 1}}\left( {dk:x \ne \pm 1;x \ne \pm 2} \right)\\
\Leftrightarrow \frac{1}{{x - 1}} - \frac{1}{{x + 1}} = \frac{1}{{x + 2}} - \frac{1}{{x - 2}}\\
\Leftrightarrow \frac{{x + 1 - x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{x - 2 - x - 2}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
\Leftrightarrow \frac{2}{{{x^2} - 1}} = \frac{{ - 4}}{{{x^2} - 4}}\\
\Leftrightarrow {x^2} - 4 = - 2\left( {{x^2} - 1} \right)\\
\Leftrightarrow {x^2} - 4 = - 2{x^2} + 2\\
\Leftrightarrow 3{x^2} = 6\\
\Leftrightarrow {x^2} = 2\\
\Leftrightarrow x = \pm \sqrt 2 \left( {tm} \right)
\end{array}$
