Đáp án:
$\begin{array}{l}
A = \dfrac{{2x + 1}}{{{x^2} + 2x + 2}}\\
\Rightarrow A.{x^2} + 2.A.x + 2A = 2x + 1\\
\Rightarrow A.{x^2} + 2.\left( {A - 1} \right).x + 2A - 1 = 0\\
+ Khi:A = 0 \Rightarrow x = - \dfrac{1}{2}\\
+ Khi:A \ne 0\\
\Delta ' \ge 0\\
\Rightarrow {\left( {A - 1} \right)^2} - A.\left( {2A - 1} \right) \ge 0\\
\Rightarrow {A^2} - 2A + 1 - 2{A^2} + A \ge 0\\
\Rightarrow {A^2} + A - 1 \le 0\\
\Rightarrow {A^2} + 2.A.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{5}{4} \le 0\\
\Rightarrow {\left( {A + \dfrac{1}{2}} \right)^2} \le \dfrac{5}{4}\\
\Rightarrow \dfrac{{ - \sqrt 5 - 1}}{2} \le A \le \dfrac{{\sqrt 5 - 1}}{2}\\
\Rightarrow \left\{ \begin{array}{l}
GTLN:A = \dfrac{{\sqrt 5 - 1}}{2}\\
GTNN:A = \dfrac{{ - \sqrt 5 - 1}}{2}
\end{array} \right.
\end{array}$
