Đáp án:
$\begin{array}{l}
B1)\\
a)A = {\left( {{x^2} - 9} \right)^2} + \left| {y - 2} \right| - 1\\
Do:\left\{ \begin{array}{l}
{\left( {{x^2} - 9} \right)^2} \ge 0\\
\left| {y - 2} \right| \ge 0
\end{array} \right.\\
\Rightarrow {\left( {{x^2} - 9} \right)^2} + \left| {y - 2} \right| \ge 0\\
\Rightarrow {\left( {{x^2} - 9} \right)^2} + \left| {y - 2} \right| - 1 \ge - 1\\
\Rightarrow A \ge - 1\\
\Rightarrow GTNN:A = - 1\,khi:\left\{ \begin{array}{l}
x = \pm 3\\
y = 2
\end{array} \right.\\
b)B = {x^4} + 3{x^2} + 2\\
Do:{x^2};{x^4} \ge 0\\
\Rightarrow {x^4} + 3{x^2} \ge 0\\
\Rightarrow {x^4} + 3{x^2} + 2 \ge 2\\
\Rightarrow B \ge 2\\
\Rightarrow GTNN:B = 2\,khi:x = 0\\
c)C = {x^2} + 4x + 100\\
= {x^2} + 4x + 4 + 96\\
= {\left( {x + 2} \right)^2} + 96 \ge 96\\
\Rightarrow GTNN:C = 96\,khi:x = - 2\\
d)D = \dfrac{{ - 1}}{{{{\left( {x + 3} \right)}^2} + 2}}\\
Do:{\left( {x + 3} \right)^2} + 2 \ge 2\\
\Rightarrow \dfrac{1}{{{{\left( {x + 3} \right)}^2} + 2}} \le \dfrac{1}{2}\\
\Rightarrow \dfrac{{ - 1}}{{{{\left( {x + 3} \right)}^2} + 2}} \ge - \dfrac{1}{2}\\
\Rightarrow D \ge - \dfrac{1}{2}\\
\Rightarrow GTNN:D = - \dfrac{1}{2}\,khi:x = - 3\\
B2)\\
a)A = - 3{x^2} - 5\left| {y - 1} \right| + 3\\
Do: - 3{x^2} - 5\left| {y - 1} \right| \le 0\\
\Rightarrow - 3{x^2} - 5\left| {y - 1} \right| + 3 \le 3\\
\Rightarrow A \le 3\\
\Rightarrow GTLN:A = 3\,khi:\left\{ \begin{array}{l}
x = 0\\
y = 1
\end{array} \right.\\
b)B = \dfrac{{2005}}{{{{\left( {x + 3} \right)}^2} + \left| {y - 1} \right| + 5}}\\
{\left( {x + 3} \right)^2} + \left| {y - 1} \right| + 5 \ge 5\\
\Rightarrow \dfrac{1}{{{{\left( {x + 3} \right)}^2} + \left| {y - 1} \right| + 5}} \le \dfrac{1}{5}\\
\Rightarrow \dfrac{{2005}}{{{{\left( {x + 3} \right)}^2} + \left| {y - 1} \right| + 5}} \le \dfrac{{2005}}{5} = 401\\
\Rightarrow B \le 401\\
\Rightarrow GTLN:B = 401\,khi:\left\{ \begin{array}{l}
x = - 3\\
y = 1
\end{array} \right.
\end{array}$
