Đáp án:
$\begin{array}{l}
B = \left| {{x^2} - 1} \right| + {\left( {x - 1} \right)^2} + {y^2}\\
Do:\left\{ \begin{array}{l}
\left| {{x^2} - 1} \right| \ge 0\\
{\left( {x - 1} \right)^2} \ge 0\\
{y^2} \ge 0
\end{array} \right.\\
\Rightarrow \left| {{x^2} - 1} \right| + {\left( {x - 1} \right)^2} + {y^2} \ge 0\\
\Rightarrow B \ge 0\\
\Rightarrow GTNN:B = 0\\
Khi:\left\{ \begin{array}{l}
{x^2} - 1 = 0\\
x - 1 = 0\\
y = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 1\\
y = 0
\end{array} \right.\\
C = \dfrac{1}{{2{{\left( {x + 1} \right)}^2} + 1}}\\
Do:{\left( {x + 1} \right)^2} \ge 0\\
\Rightarrow 2{\left( {x + 1} \right)^2} \ge 0\\
\Rightarrow 2{\left( {x + 1} \right)^2} + 1 \ge 1\\
\Rightarrow \dfrac{1}{{2{{\left( {x + 1} \right)}^2} + 1}} \le 1\\
\Rightarrow C \le 1\\
\Rightarrow GTLN:C = 1\,khi:x = - 1\\
D = {\left( {2x - 3} \right)^2} + {\left( {y - \dfrac{1}{2}} \right)^2} + 2021 \ge 2021\\
\Rightarrow GTNN:D = 2021\\
Khi:\left\{ \begin{array}{l}
x = \dfrac{3}{2}\\
y = \dfrac{1}{2}
\end{array} \right.\\
E = 2{\left( {2x + 1} \right)^2} + \left| { - 3\left( {{x^2} - 1} \right)} \right| \ge 2.{\left( { - 2 + 1} \right)^2} = 2\\
\Rightarrow GTNN:E = 2\\
Khi:x = - 1\\
F = \dfrac{{ - 1}}{{2{{\left( {x + 1} \right)}^2} + 1}} \ge - 1\\
\Rightarrow GTNN:F = - 1\\
Khi:x = - 1
\end{array}$
