Chúc bn học tốt. 10 điểm nhé

Đáp án:

$\begin{array}{l}
a)Dkxd:x \ne 2;x \ne  - 2\\
\dfrac{{x + 1}}{{x - 2}} - \dfrac{5}{{x + 2}} = \dfrac{{12}}{{{x^2} - 4}} + 1\\
 \Leftrightarrow \dfrac{{\left( {x + 1} \right)\left( {x + 2} \right) - 5.\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{12 + {x^2} - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 \Leftrightarrow {x^2} + 3x + 2 - 5x + 10 = {x^2} + 8\\
 \Leftrightarrow  - 2x + 12 = 8\\
 \Leftrightarrow  - 2x =  - 4\\
 \Leftrightarrow x = 2\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
b)\left| {2x + 6} \right| - x = 3\\
 \Leftrightarrow \left| {2x + 6} \right| = x + 3\\
 \Leftrightarrow 2.\left| {x + 3} \right| = x + 3\left( {dk:x \ge  - 3} \right)\\
 \Leftrightarrow 2.\left( {x + 3} \right) = x + 3\\
 \Leftrightarrow x + 3 = 0\\
 \Leftrightarrow x =  - 3\\
Vậy\,x =  - 3\\
c)\dfrac{{2x + 2}}{5} + \dfrac{3}{{10}} < \dfrac{{3x - 2}}{4}\\
 \Leftrightarrow \dfrac{{4.\left( {2x + 2} \right) + 3.2}}{{20}} < \dfrac{{5.\left( {3x - 2} \right)}}{{20}}\\
 \Leftrightarrow 8x + 8 + 6 < 15x - 10\\
 \Leftrightarrow 7x > 24\\
 \Leftrightarrow x > \dfrac{{24}}{7}\\
Vậy\,x > \dfrac{{24}}{7}
\end{array}$