Đáp án:
$\begin{array}{l}
P = {\left( {5 - \left| {x + 1} \right|} \right)^2} - 2\left| {y - \dfrac{1}{3}} \right|\\
Do:\left| {x + 1} \right| \ge 0\\
\Rightarrow - \left| {x + 1} \right| \le 0\\
\Rightarrow 5 - \left| {x + 1} \right| \le 5\\
\Rightarrow {\left( {5 - \left| {x + 1} \right|} \right)^2} \le 25\\
Do:\left| {y - \dfrac{1}{3}} \right| \ge 0\\
\Rightarrow - 2\left| {y - \dfrac{1}{3}} \right| \le 0\\
\Rightarrow {\left( {5 - \left| {x + 1} \right|} \right)^2} - 2\left| {y - \dfrac{1}{3}} \right| \le 25\\
\Rightarrow P \le 25\\
\Rightarrow GTLN:P = 25\\
Khi:\left\{ \begin{array}{l}
x + 1 = 0\\
y - \dfrac{1}{3} = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = - 1\\
y = \dfrac{1}{3}
\end{array} \right.
\end{array}$
