Đáp án:
$\begin{array}{l}
a)P = 3\left| {2x + 5} \right| - 7\\
Do:3.\left| {2x + 5} \right| \ge 0\\
\Leftrightarrow 3\left| {2x + 5} \right| - 7 \ge - 7\\
\Leftrightarrow P \ge - 7\\
\Leftrightarrow GTNN:P = - 7\,khi:x = - \dfrac{5}{2}\\
b)Q = \left| {x - 3} \right| + \left| {x - 5} \right|\\
= \left| {x - 3} \right| + \left| {5 - x} \right|\\
\ge \left| {x - 3 + 5 - x} \right|\\
\Leftrightarrow Q \ge 2\\
\Leftrightarrow GTNN:Q = 2\,khi:3 \le x \le 5\\
c)R = {\left( {2x - 3} \right)^2} - 14 \ge - 14\\
\Leftrightarrow R \ge - 14\\
\Leftrightarrow GTNN:R = - 14\,khi:x = \dfrac{3}{2}\\
d)H = {\left( {2x - y} \right)^2} + \left| {x - 3} \right| + 7\\
Do:\left\{ \begin{array}{l}
{\left( {2x - y} \right)^2} \ge 0\\
\left| {x - 3} \right| \ge 0
\end{array} \right.\\
\Leftrightarrow {\left( {2x - y} \right)^2} + \left| {x - 3} \right| \ge 0\\
\Leftrightarrow {\left( {2x - y} \right)^2} + \left| {x - 3} \right| + 7 \ge 7\\
\Leftrightarrow H \ge 7\\
\Leftrightarrow GTNN:H = 7\,\\
khi:\left\{ \begin{array}{l}
2x = y\\
x = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y = 6\\
x = 3
\end{array} \right.
\end{array}$
