Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
A = - {x^2} + 4x + 4 = - \left( {{x^2} - 4x + 4} \right) + 8 = 8 - {\left( {x - 2} \right)^2} \le 8,\,\,\,\forall x\\
\Rightarrow {A_{\max }} = 8 \Leftrightarrow {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x = 2\\
B = 4 - 16{x^2} - 8x = - \left( {16{x^2} + 8x + 1} \right) + 5 = 5 - {\left( {4x + 1} \right)^2} \le 5,\,\,\,\forall x\\
\Rightarrow {B_{\max }} = 5 \Leftrightarrow {\left( {4x + 1} \right)^2} = 0 \Leftrightarrow x = - \frac{1}{4}\\
2,\\
{x^2} + 2x + {y^2} - 6y + 10 = 0\\
\Leftrightarrow \left( {{x^2} + 2x + 1} \right) + \left( {{y^2} - 6y + 9} \right) = 0\\
\Leftrightarrow {\left( {x + 1} \right)^2} + {\left( {y - 3} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 1} \right)^2} = 0\\
{\left( {y - 3} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 1\\
y = 3
\end{array} \right.\\
3,\\
A = \left( {4 + 1} \right)\left( {{4^2} + 1} \right)\left( {{4^4} + 1} \right)\left( {{4^8} + 1} \right)\left( {{4^{16}} + 1} \right)\\
\Leftrightarrow 3A = 3.\left( {4 + 1} \right)\left( {{4^2} + 1} \right)\left( {{4^4} + 1} \right)\left( {{4^8} + 1} \right)\left( {{4^{16}} + 1} \right)\\
\Leftrightarrow 3A = \left( {4 - 1} \right).\left( {4 + 1} \right)\left( {{4^2} + 1} \right)\left( {{4^4} + 1} \right)\left( {{4^8} + 1} \right)\left( {{4^{16}} + 1} \right)\\
\Leftrightarrow 3A = \left( {{4^2} - 1} \right).\left( {{4^2} + 1} \right)\left( {{4^4} + 1} \right)\left( {{4^8} + 1} \right)\left( {{4^{16}} + 1} \right)\\
\Leftrightarrow 3A = \left( {{4^4} - 1} \right).\left( {{4^4} + 1} \right)\left( {{4^8} + 1} \right)\left( {{4^{16}} + 1} \right)\\
\Leftrightarrow 3A = \left( {{4^8} - 1} \right)\left( {{4^8} + 1} \right)\left( {{4^{16}} + 1} \right)\\
\Leftrightarrow 3A = \left( {{4^{16}} - 1} \right)\left( {{4^{16}} + 1} \right)\\
\Leftrightarrow 3A = {4^{32}} - 1\\
\Leftrightarrow 3A = B
\end{array}\)
