Đáp án:
$\begin{array}{l}
y = \dfrac{{{x^2} - 2x + 2}}{{{x^2} + 2x + 2}}\\
\Leftrightarrow y.{x^2} + 2y.x + 2y = {x^2} - 2x + 2\\
\Leftrightarrow \left( {y - 1} \right).{x^2} + \left( {2y + 2} \right).x + 2y - 2 = 0\\
\Leftrightarrow \left( {y - 1} \right).{x^2} + 2.\left( {y + 1} \right).x + 2y - 2 = 0\\
\Leftrightarrow \Delta ' \ge 0\\
\Leftrightarrow {\left( {y + 1} \right)^2} - \left( {y - 1} \right).\left( {2y - 2} \right) \ge 0\\
\Leftrightarrow {y^2} + 2y + 1 - 2{\left( {y - 1} \right)^2} \ge 0\\
\Leftrightarrow {y^2} + 2y + 1 - 2{y^2} + 4y - 2 \ge 0\\
\Leftrightarrow {y^2} - 6y + 1 \le 0\\
\Leftrightarrow {\left( {y - 3} \right)^2} - 8 \le 0\\
\Leftrightarrow {\left( {y - 3} \right)^2} \le 8\\
\Leftrightarrow - 2\sqrt 2 \le y - 3 \le 2\sqrt 2 \\
\Leftrightarrow 3 - 2\sqrt 2 \le y \le 3 + 2\sqrt 2 \\
\Leftrightarrow \left\{ \begin{array}{l}
GTNN:y = 3 - 2\sqrt 2 \\
GTLN:y = 3 + 2\sqrt 2
\end{array} \right.
\end{array}$
