Đáp án:
$\begin{array}{l}
a)Dkxd:a \ne \dfrac{5}{2};a \ne \dfrac{2}{3}\\
\dfrac{{2a - 9}}{{2a - 5}} + \dfrac{{3a}}{{3a - 2}} = 2\\
\Rightarrow \dfrac{{\left( {2a - 9} \right)\left( {3a - 2} \right) + 3a\left( {2a - 5} \right)}}{{\left( {2a - 5} \right)\left( {3a - 2} \right)}} = 2\\
\Rightarrow 6{a^2} - 4a - 27a + 18 + 6{a^2} - 15a\\
= 2\left( {6{a^2} - 4a - 15a + 10} \right)\\
\Rightarrow 12{a^2} - 46a + 18 = 12{a^2} - 38a + 20\\
\Rightarrow 8a = - 2\\
\Rightarrow a = - \dfrac{1}{4}\left( {tmdk} \right)\\
Vậy\,a = - \dfrac{1}{4}\\
b)Dkxd:a \ne - \dfrac{4}{3};a \ne - 4\\
\dfrac{{3a + 2}}{{3a + 4}} + \dfrac{{a - 2}}{{a + 4}} = 2\\
\Rightarrow \dfrac{{\left( {3a + 2} \right)\left( {a + 4} \right) + \left( {a - 2} \right)\left( {3a + 4} \right)}}{{\left( {3a + 4} \right)\left( {a + 4} \right)}} = 2\\
\Rightarrow 3{a^2} + 14a + 8 + 3{a^2} - 2a - 8\\
= 2\left( {3{a^2} + 16a + 16} \right)\\
\Rightarrow 6{a^2} + 12a = 6{a^2} + 32a + 32\\
\Rightarrow 20a = - 32\\
\Rightarrow a = - \dfrac{8}{5}\left( {tmdk} \right)\\
Vậy\,a = - \dfrac{8}{5}
\end{array}$
