Đáp án:
$\begin{array}{l}
3)a){\left( {6x + 1} \right)^2} + {\left( {6x - 1} \right)^2} - 2\left( {1 + 6x} \right)\left( {6x - 1} \right)\\
= {\left( {6x - 1 - 6x - 1} \right)^2}\\
= 4\\
b)3\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
= \left( {{2^2} - 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
= \left( {{2^4} - 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
= \left( {{2^8} - 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
= \left( {{2^{16}} - 1} \right)\left( {{2^{16}} + 1} \right)\\
= {2^{32}} - 1\\
4)C = {x^2} - 4xy + 5{y^2} + 10x - 22y + 28\\
= {x^2} + 4{y^2} + 25 + 10x - 4xy - 20y + {y^2} - 2y + 1 + 2\\
= {\left( {x - 2y + 5} \right)^2} + {\left( {y - 1} \right)^2} + 2 \ge 2\forall x,y\\
\Rightarrow GTNN:C = 2 \Leftrightarrow \left\{ \begin{array}{l}
x - 2y + 5 = 0\\
y = 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = - 3\\
y = 1
\end{array} \right.\\
5)B = - {x^2} + 6x - 11\\
= - \left( {{x^2} - 6x + 9} \right) - 2\\
= - {\left( {x - 3} \right)^2} - 2 \le - 2\forall x\\
\Rightarrow GTLN:B = - 2 \Leftrightarrow x = 3
\end{array}$
