Đáp án:
$\begin{array}{l}
c)\frac{2}{{x + y}} + \frac{1}{{x - y}} + \frac{{3x}}{{{y^2} - {x^2}}} = \frac{{2\left( {x - y} \right) + x + y - 3x}}{{\left( {x + y} \right)\left( {x - y} \right)}}\\
= \frac{{2x - 2y + x + y - 3x}}{{\left( {x + y} \right)\left( {x - y} \right)}} = \frac{{ - y}}{{{x^2} - {y^2}}} = \frac{y}{{{y^2} - {x^2}}}\\
d)\frac{{x + 1}}{{2x - 2}} + \frac{{3 - {x^2}}}{{3{x^2} - 3}} = \frac{{x + 1}}{{2\left( {x - 1} \right)}} + \frac{{3 - {x^2}}}{{3\left( {{x^2} - 1} \right)}}\\
= \frac{{\left( {x + 1} \right)3\left( {x + 1} \right) + 2\left( {3 - {x^2}} \right)}}{{6\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{3{x^2} + 6x + 3 + 6 - 2{x^2}}}{{6\left( {x + 1} \right)\left( {x - 1} \right)}}\\
= \frac{{{x^2} + 6x + 9}}{{6\left( {{x^2} - 1} \right)}} = \frac{{{{\left( {x + 3} \right)}^2}}}{{6\left( {{x^2} - 1} \right)}}
\end{array}$
em tách những ý còn lại ra hỏi thành câu khác nhé
