Đáp án:
$\begin{array}{l}
a)\left| {2x + 1} \right| = x + 2\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 1 = x + 2\\
2x + 1 = - x - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x - x = 2 - 1\\
2x + x = - 2 - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
3x = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.\\
Vậy\,x = 1;x = - 1\\
b){x^2} - 2x + 8 = \left| {{x^2} - 1} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 2x + 8 = {x^2} - 1\\
{x^2} - 2x + 8 = - {x^2} + 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 8\\
2{x^2} - 2x + 7 = 0\left( {vn} \right)
\end{array} \right.\\
\Leftrightarrow x = 4\\
Vậy\,x = 4\\
c)Dkxd:x\# \dfrac{3}{2};x\# - 1\\
\dfrac{{x - 1}}{{2x - 3}} = \dfrac{{ - 3x + 1}}{{\left| {x + 1} \right|}}\\
+ Khi:x > - 1\\
\Leftrightarrow \dfrac{{x - 1}}{{2x - 3}} = \dfrac{{ - 3x + 1}}{{x + 1}}\\
\Leftrightarrow \left( {x - 1} \right)\left( {x + 1} \right) = \left( {2x - 3} \right)\left( { - 3x + 1} \right)\\
\Leftrightarrow {x^2} - 1 = - 6{x^2} + 11x - 3\\
\Leftrightarrow 7{x^2} - 11x + 2 = 0\\
\Leftrightarrow x = \dfrac{{11 \pm \sqrt {65} }}{{14}}\left( {tm} \right)\\
+ Khi:x < - 1\\
\Leftrightarrow \dfrac{{x - 1}}{{2x - 3}} = \dfrac{{ - 3x + 1}}{{ - \left( {x + 1} \right)}}\\
\Leftrightarrow - \left( {x - 1} \right)\left( {x + 1} \right) = \left( {2x - 3} \right)\left( { - 3x + 1} \right)\\
\Leftrightarrow 1 - {x^2} = - 6{x^2} + 11x - 3\\
\Leftrightarrow 5{x^2} - 11x + 4 = 0\\
\Leftrightarrow x = \dfrac{{11 \pm \sqrt {41} }}{{10}}\left( {ktm} \right)\\
Vậy\,x = \dfrac{{11 \pm \sqrt {65} }}{{14}}\\
d)\left| {5x + 2} \right| + \left| {3x - 4} \right| = 4x + 5\\
+ Khi:x \ge \dfrac{4}{3}\\
\Leftrightarrow 5x + 2 + 3x - 4 = 4x + 5\\
\Leftrightarrow 4x = 7\\
\Leftrightarrow x = \dfrac{7}{4}\left( {tm} \right)\\
+ Khi:\dfrac{{ - 2}}{5} \le x < \dfrac{4}{3}\\
\Leftrightarrow 5x + 2 + 4 - 3x = 4x + 5\\
\Leftrightarrow 2x = 2\\
\Leftrightarrow x = 1\left( {tm} \right)\\
+ Khi:x < \dfrac{{ - 2}}{5}\\
\Leftrightarrow - 5x - 2 + 4 - 3x = 4x + 5\\
\Leftrightarrow 12x = - 3\\
\Leftrightarrow x = - \dfrac{1}{4}\left( {ktm} \right)\\
Vậy\,x = 1;x = \dfrac{7}{4}
\end{array}$
