Giúp em câu D vs ạ :33

Đáp án:

$\begin{array}{l}
 + ){x^2} + 2x + 1 = {\left( {x + 1} \right)^2}\\
 + )2{x^2} - 2 = 2.\left( {{x^2} - 1} \right) = 2.\left( {x - 1} \right)\left( {x + 1} \right)\\
 \Leftrightarrow MSC = 6{x^2}{\left( {x + 1} \right)^2}.\left( {x - 1} \right)\\
 + \dfrac{2}{{3{x^2}}} = \dfrac{{2.2{{\left( {x + 1} \right)}^2}.\left( {x - 1} \right)}}{{6{x^2}{{\left( {x + 1} \right)}^2}.\left( {x - 1} \right)}}\\
 = \dfrac{{4\left( {x + 1} \right)\left( {{x^2} - 1} \right)}}{{6{x^2}{{\left( {x + 1} \right)}^2}.\left( {x - 1} \right)}}\\
 = \dfrac{{4{x^3} + 4{x^2} - 4x - 4}}{{6{x^2}{{\left( {x + 1} \right)}^2}.\left( {x - 1} \right)}}\\
 + )\dfrac{x}{{{x^2} + 2x + 1}} = \dfrac{{x.6{x^2}\left( {x - 1} \right)}}{{6{x^2}{{\left( {x + 1} \right)}^2}.\left( {x - 1} \right)}}\\
 = \dfrac{{6{x^4} - 6{x^3}}}{{6{x^2}{{\left( {x + 1} \right)}^2}.\left( {x - 1} \right)}}\\
 + )\dfrac{y}{{2{x^2} - 2}} = \dfrac{{y.3{x^2}\left( {x + 1} \right)}}{{6{x^2}{{\left( {x + 1} \right)}^2}.\left( {x - 1} \right)}}\\
 = \dfrac{{3{x^3}y + 3{x^2}y}}{{6{x^2}{{\left( {x + 1} \right)}^2}.\left( {x - 1} \right)}}
\end{array}$