Đáp án:
$\begin{array}{l}
a)\,\,2\\
b)\,\,\,x - 1.
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
a)\,\,\,\frac{3}{{x + 3}} - \frac{{x - 6}}{{{x^2} + 3x}} = \frac{3}{{x + 3}} - \frac{{x - 6}}{{x\left( {x + 3} \right)}}\\
= \frac{{3x - x + 6}}{{x + 3}} = \frac{{2x + 6}}{{x + 3}} = \frac{{2\left( {x + 3} \right)}}{{x + 3}} = 2.\\
b)\,\,\frac{{2{x^2} - x}}{{x - 1}} + \frac{{x + 1}}{{1 - x}} + \frac{{2 - {x^2}}}{{x - 1}}\\
= \frac{{2{x^2} - x}}{{x - 1}} - \frac{{x + 1}}{{x - 1}} + \frac{{2 - {x^2}}}{{x - 1}}\\
= \frac{{2{x^2} - x - x - 1 + 2 - {x^2}}}{{x - 1}}\\
= \frac{{{x^2} - 2x + 1}}{{x - 1}} = \frac{{{{\left( {x - 1} \right)}^2}}}{{x - 1}} = x - 1.
\end{array}$
