Đáp án:
$\begin{array}{l}
1)A = \sqrt {16{x^2} + 8x + 1} \\
= \sqrt {{{\left( {4x} \right)}^2} + 2.4x.1 + 1} \\
= \sqrt {{{\left( {4x + 1} \right)}^2}} \\
= \left| {4x + 1} \right|\\
Thay\,x = 2\\
\Rightarrow A = \left| {4.2 + 1} \right| = \left| 9 \right| = 9\\
b)A = 5\\
\Rightarrow \left| {4x + 1} \right| = 5\\
\Rightarrow \left[ \begin{array}{l}
4x + 1 = 5\\
4x + 1 = - 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
4x = 4\\
4x = - 6
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = - \frac{3}{2}
\end{array} \right.\\
Vay\,x = 1;x = \frac{{ - 3}}{2}\\
B2)Khi:x = \sqrt {15} \\
B = \sqrt {15{x^2} + 8x.\sqrt {15} + 15} \\
= \sqrt {15.{{\left( {\sqrt {15} } \right)}^2} + 8.\sqrt {15} .\sqrt {15} + 15} \\
= \sqrt {15.15 + 8.15 + 15} \\
= \sqrt {15.\left( {15 + 8 + 1} \right)} \\
= \sqrt {15.24} \\
= \sqrt {3.5.4.6} \\
= 2.3\sqrt {5.2} \\
= 6\sqrt {10}
\end{array}$
