Đáp án:
$\begin{array}{l}
Dkxd:x \ne \frac{3}{2}\\
a)\left| {x + \frac{1}{3}} \right| - 4 = - 1\\
\Rightarrow \left| {x + \frac{1}{3}} \right| = 3\\
\Rightarrow \left[ \begin{array}{l}
x + \frac{1}{3} = 3\\
x + \frac{1}{3} = - 3
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \frac{8}{3}\left( {tm} \right)\\
x = - \frac{{10}}{3}\left( {tm} \right)
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
A = \frac{{32}}{7}\\
A = \frac{{94}}{{29}}
\end{array} \right.\\
b)A = \frac{{7x - 8}}{{2x - 3}}\\
= \frac{{\frac{7}{2}.\left( {2x - 3} \right) + \frac{{21}}{2} - 8}}{{2x - 3}}\\
= \frac{7}{2} + \frac{5}{{4x - 6}}\\
{A_{\max }} \Rightarrow \frac{5}{{4x - 6}}\left( {max} \right)\\
\Rightarrow {\left( {4x - 6} \right)_{\min }}\\
\Rightarrow 4x - 6 = 1\\
\Rightarrow 4x = 7\\
\Rightarrow x = \frac{7}{4}
\end{array}$
