Đáp án:
$\begin{array}{l}
a)\left( {\frac{1}{{{x^2} + x - 2}} - \frac{x}{{x + 1}}} \right):\left( {\frac{1}{{{x^2} + x - 2}}} \right)\\
= \left( {\frac{1}{{\left( {x - 1} \right)\left( {x + 2} \right)}} - \frac{x}{{x + 1}}} \right).\left( {x - 1} \right)\left( {x + 2} \right)\\
= \frac{{x + 1 - x\left( {{x^2} + x - 2} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)\left( {x + 1} \right)}}.\left( {x - 1} \right)\left( {x + 2} \right)\\
= \frac{{ - {x^3} - {x^2} + 3x + 1}}{{x + 1}}\\
b)\left( {\frac{{3x}}{{1 - 3x}} + \frac{{2x}}{{3x + 1}}} \right):\frac{{6{x^2} + 10x}}{{1 - 6x + 9{x^2}}}\\
= \frac{{3x\left( {3x + 1} \right) + 2x\left( {1 - 3x} \right)}}{{\left( {1 - 3x} \right)\left( {3x + 1} \right)}}.\frac{{{{\left( {3x - 1} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
= \frac{{3{x^2} + 5x}}{{\left( {3x + 1} \right).2x\left( {3x + 5} \right)}}.\left( {1 - 3x} \right)\\
= \frac{{1 - 3x}}{{6x + 2}}
\end{array}$
