Giải thích các bước giải:
\(
\begin{array}{l}
A = {\rm{[}}\frac{{{\rm{3(a + 2)}}}}{{{\rm{a}}^{\rm{3}} + a^2 + a + 1}} + \frac{{2a^2 - a - 10}}{{a^3 - a^2 + a - 1}}{\rm{]:[}}\frac{{\rm{5}}}{{{\rm{a}}^{\rm{2}} + 1}} + \frac{3}{{2a + 2}} - \frac{3}{{2a - 2}}{\rm{]}} \\
{\rm{a)Đk:a}} \ne \pm {\rm{1}} \\
{\rm{A = [}}\frac{{3(a + 2)}}{{(a + 1)(a^2 + 1)}} + \frac{{2a^2 - a - 10}}{{(a - 1)(a^2 + 1)}}{\rm{]:[}}\frac{{\rm{5}}}{{{\rm{a}}^{\rm{2}} + 1}} + \frac{3}{{2(a + 1)}} - \frac{3}{{2(a - 1)}}{\rm{]}} \\
{\rm{ = }}\frac{{{\rm{3(a + 2)(a - 1) + (2a}}^{\rm{2}} - a - 10)(a + 1)}}{{(a - 1)(a + 1)(a^2 + 1)}}:\frac{{10(a + 1)(a - 1) + 3(a^2 + 1)(a - 1) - 3(a^2 + 1)(a + 1)}}{{2(a - 1)(a + 1)(a^2 + 1)}} \\
= \frac{{2a^3 + 4a^2 - 8a - 16}}{{(a - 1)(a + 1)(a^2 + 1)}}.\frac{{2(a - 1)(a + 1)(a^2 + 1)}}{{4a^2 - 4}} \\
= \frac{{a^3 + 2a - 4a - 8}}{{a^2 - 1}} = \frac{{(a + 2)(a - 4)}}{{a^2 - 1}} \\
b)a = 2 \Leftrightarrow A = \frac{{(2 + 2)(2 - 4)}}{{2^2 - 1}} = \frac{{ - 8}}{3} \\
c)A = 0 \Rightarrow \frac{{(a + 2)(a - 4)}}{{a^2 - 1}} = 0(a \ne \pm 1) \\
\Leftrightarrow (a + 2)(a - 4) = 0 \\
\Leftrightarrow \left[ {\begin{array}{*{20}c}
{a = - 2} \\
{a = 4} \\
\end{array}} \right. \\
\end{array}
\)
