Đáp án:
\(\frac{{25000}}{{3\pi }} \le f \le \frac{{250000}}{{3\pi }}\)
Giải thích các bước giải:
\(\begin{array}{l}
{f_1} = \frac{1}{{2\pi \sqrt {L{C_1}} }} = \frac{1}{{2\pi \sqrt {{{3.10}^{ - 3}}{{.12.10}^{ - 9}}} }} = \frac{{250000}}{{3\pi }}\\
{f_2} = \frac{1}{{2\pi \sqrt {L{C_2}} }} = \frac{1}{{2\pi \sqrt {{{3.10}^{ - 3}}{{.1200.10}^{ - 9}}} }} = \frac{{25000}}{{3\pi }}\\
{f_2} \le f \le {f_1}\\
\frac{{25000}}{{3\pi }} \le f \le \frac{{250000}}{{3\pi }}
\end{array}\)