Đáp án:
$\begin{array}{l}
F = \frac{1}{x}\\
\Rightarrow B = \frac{{{x^2} + 9}}{x} - 5\\
\Rightarrow B + 5 = \frac{{{x^2} + 9}}{x}\\
\Rightarrow B.x + 5.x = {x^2} + 9\\
\Rightarrow {x^2} - \left( {B + 5} \right).x + 9 = 0\\
\Rightarrow \Delta \ge 0\\
\Rightarrow {\left( {B + 5} \right)^2} - 36 \ge 0\\
\Rightarrow \left[ \begin{array}{l}
B + 5 \ge 6\\
B + 5 \le - 6
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
B \ge 1\\
B \le - 11
\end{array} \right.\\
\Rightarrow {B_{\min }} = 1\,khi:x = 3
\end{array}$
