Đáp án:
$\begin{array}{l}
a)A = 3{x^2} + 1\\
Do:{x^2} \ge 0\\
\Leftrightarrow 3{x^2} + 1 \ge 1\\
\Leftrightarrow A \ge 1\\
\Leftrightarrow GTNN:A = 1\,khi:x = 0\\
B = \left| {x + 1} \right| + 2{\left( {6 - 3y} \right)^2} + 3\\
DO:\left\{ \begin{array}{l}
\left| {x + 1} \right| \ge 0\\
2{\left( {6 - 3y} \right)^2} \ge 0
\end{array} \right.\\
\Leftrightarrow \left| {x + 1} \right| + 2{\left( {6 - 3y} \right)^2} + 3 \ge 3\\
\Leftrightarrow B \ge 3\\
\Leftrightarrow GTNN:B = 3\\
Khi:\left\{ \begin{array}{l}
x = - 1\\
y = 2
\end{array} \right.\\
b)C = 8 - 6\left| {x - 7} \right| - {\left( {{y^2} - 16} \right)^2}\\
Do: - 6\left| {x - 7} \right| - \left( {{y^2} - 16} \right) \le 0\\
\Leftrightarrow 8 - 6\left| {x - 7} \right| - {\left( {{y^2} - 16} \right)^2} \le 8\\
\Leftrightarrow C \le 8\\
\Leftrightarrow GTLN:C = 8\,khi:\left\{ \begin{array}{l}
x = 7\\
y = \pm 4
\end{array} \right.\\
c){x^2} + {\left( {y - 1} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
{x^2} \ge 0\\
{\left( {y - 1} \right)^2} \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ {{x^2} + {{\left( {y - 1} \right)}^2} \ge 0} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} = 0\\
y - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
y = 1
\end{array} \right.\\
Vậy\,x = 0;y = 1
\end{array}$
