Đáp án:

 60m và 80m 

Giải thích các bước giải:

* Khi C1//C2:

\(\lambda _1^2 = {\left( {\dfrac{c}{{{f_1}}}} \right)^2} = \dfrac{{4{\pi ^2}{c^2}}}{{\dfrac{1}{{L\left( {{C_1} + {C_2}} \right)}}}} = 4{\pi ^2}{c^2}L\left( {{C_1} + {C_2}} \right)\)

* Khi C1 nt C2:

\(\lambda _2^2 = {\left( {\dfrac{c}{{{f_2}}}} \right)^2} = \dfrac{{4{\pi ^2}{c^2}}}{{\dfrac{1}{{L\left( {\dfrac{{{C_1}.{C_2}}}{{{C_1} + {C_2}}}} \right)}}}} = 4{\pi ^2}{c^2}L\left( {\dfrac{{{C_1}.{C_2}}}{{{C_1} + {C_2}}}} \right)\)

Suy ra:

\(\begin{array}{l}
\dfrac{{\lambda _1^2}}{{\lambda _2^2}} = \dfrac{{{C_1} + {C_2}}}{{\dfrac{{{C_1}.{C_2}}}{{{C_1} + {C_2}}}}} \Rightarrow \dfrac{{625}}{{144}} = \dfrac{{C_1^2 + 2{C_1}.{C_2} + C_2^2}}{{{C_1}.{C_2}}}\\
 \Rightarrow \left( \begin{array}{l}
{C_1} = \dfrac{{16}}{9}{C_2}\\
{C_1} = \dfrac{9}{{16}}{C_2}
\end{array} \right.
\end{array}\)

2 trường hợp tương tự nhau nên ta chỉ xét trường hợp 1.

* Khi chỉ dùng C1:

\(\lambda _3^2 = {\left( {\dfrac{c}{{{f_3}}}} \right)^2} = \dfrac{{4{\pi ^2}{c^2}}}{{\dfrac{1}{{L{C_1}}}}} = 4{\pi ^2}{c^2}L{C_1}\)

Suy ra:

\(\dfrac{{\lambda _1^2}}{{\lambda _3^2}} = \dfrac{{{C_1} + {C_2}}}{{{C_1}}} = \dfrac{{25}}{{16}} \Rightarrow {\lambda _3} = 80m\)

* Khi chỉ dùng C2:

\(\lambda _4^2 = {\left( {\dfrac{c}{{{f_4}}}} \right)^2} = \dfrac{{4{\pi ^2}{c^2}}}{{\dfrac{1}{{L{C_2}}}}} = 4{\pi ^2}{c^2}L{C_2}\)

Suy ra:

\(\dfrac{{\lambda _1^2}}{{\lambda _4^2}} = \dfrac{{{C_1} + {C_2}}}{{{C_1}}} = \dfrac{{25}}{9} \Rightarrow {\lambda _4} = 60m\)