Đáp án:
$\begin{array}{l}
1)A = \dfrac{{12}}{{\left| x \right| - 3}}\\
Do:\left| x \right| \ge 0\\
\Rightarrow \left| x \right| - 3 \ge - 3\\
\Rightarrow \dfrac{1}{{\left| x \right| - 3}} \le \dfrac{1}{{ - 3}}\\
\Rightarrow \dfrac{{12}}{{\left| x \right| - 3}} \le \dfrac{{12}}{{ - 3}} = - 4\\
\Rightarrow A \le - 4\\
\Rightarrow GTLN:A = - 4\\
Khi:\left| x \right| = 0\\
\Rightarrow x = 0\\
2)B = x - \left| x \right|\\
+ Khi:x = 0 \Rightarrow B = 0\\
+ Khi:x > 0 \Rightarrow \left| x \right| = x\\
\Rightarrow B = x - x = 0\\
+ Khi:x < 0 \Rightarrow \left| x \right| = - x\\
\Rightarrow B = x - \left( { - x} \right) = 2x < 0\\
\Rightarrow GTLN:B = 0\,Khi:x \ge 0\\
3)\\
a)2\left| {2x - 3} \right| = \dfrac{1}{2}\\
\Rightarrow \left| {2x - 3} \right| = \dfrac{1}{4}\\
\Rightarrow \left[ \begin{array}{l}
2x - 3 = \dfrac{1}{4}\\
2x - 3 = - \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x = \dfrac{{13}}{4}\\
2x = \dfrac{{11}}{4}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{13}}{8}\\
x = \dfrac{{11}}{8}
\end{array} \right.\\
b)\left| {x + 15} \right| + 1 = 3x\\
\Rightarrow \left| {x + 15} \right| = 3x - 1\left( {dk:x \ge \dfrac{1}{3}} \right)\\
Do:x \ge \dfrac{1}{3} \Rightarrow \left| {x + 15} \right| = x + 15\\
\Rightarrow x + 15 = 3x - 1\\
\Rightarrow 3x - x = 15 + 1\\
\Rightarrow 2x = 16\\
\Rightarrow x = 8\left( {tmdk} \right)\\
Vay\,x = 8
\end{array}$
