Đáp án:
$\begin{array}{l}
a)\frac{{ - 3}}{x} + \frac{{5x}}{{{x^2}}} - \frac{5}{{x + 3}}\left( {đkxđ:x \ne 0;x \ne - 3} \right)\\
= \frac{{ - 3}}{x} + \frac{5}{x} - \frac{5}{{x + 3}}\\
= \frac{2}{x} - \frac{5}{{x + 3}}\\
= \frac{{2\left( {x + 3} \right) - 5x}}{{x\left( {x + 3} \right)}}\\
= \frac{{2x + 6 - 5x}}{{x\left( {x + 3} \right)}}\\
= \frac{{6 - 3x}}{{x\left( {x + 3} \right)}}\\
b)\frac{2}{{x + 2}} - \frac{{ - 8}}{{{x^2} - 4}} + \frac{x}{{x - 2}}\left( {đkxđ:x \ne 2;x \ne - 2} \right)\\
= \frac{2}{{x + 2}} + \frac{8}{{\left( {x + 2} \right)\left( {x - 2} \right)}} + \frac{x}{{x - 2}}\\
= \frac{{2\left( {x - 2} \right) + 8 + x\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
= \frac{{2x - 4 + 8 + {x^2} + 2x}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
= \frac{{{x^2} + 4x + 4}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
= \frac{{{{\left( {x + 2} \right)}^2}}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
= \frac{{x + 2}}{{x - 2}}
\end{array}$
