Đáp án:

a) \(\dfrac{{{x^2}}}{{x - 20}}\)

b) \(x = 40cm\)

Giải thích các bước giải:

a) Ta có:

\(\begin{array}{l}
\dfrac{{AB}}{{A'B'}} = \dfrac{{OA}}{{OA'}} = \dfrac{x}{{d'}}\\
\dfrac{{AB}}{{A'B'}} = \dfrac{{OI}}{{A'B'}} = \dfrac{{OF'}}{{OA' - OF'}} = \dfrac{{20}}{{d' - 20}}\\
 \Rightarrow \dfrac{x}{{d'}} = \dfrac{{20}}{{d' - 20}} \Rightarrow xd' - 20x = 20d'\\
 \Rightarrow d'\left( {x - 20} \right) = 20x\\
 \Rightarrow d' = \dfrac{{20x}}{{x - 20}}\\
L = x + d' = x + \dfrac{{20x}}{{x - 20}} = \dfrac{{{x^2}}}{{x - 20}}
\end{array}\)

b) Ta có:

\(\begin{array}{l}
L = \dfrac{{{x^2}}}{{x - 20}} = \dfrac{{{x^2} - {{20}^2} + {{20}^2}}}{{x - 20}} = \dfrac{{\left( {x - 20} \right)\left( {x + 20} \right) + {{20}^2}}}{{x - 20}}\\
 \Rightarrow L = x + 20 + \dfrac{{400}}{{x - 20}} = \left( {x - 20} \right) + \dfrac{{400}}{{x - 20}} + 40
\end{array}\)

Áp dụng BĐT Cô si:

\(\begin{array}{l}
L \ge 2\sqrt {\left( {x - 20} \right).\dfrac{{400}}{{x - 20}}}  + 40\\
 \Rightarrow L \ge 2\sqrt {400}  + 40 = 80cm
\end{array}\)

Dấu = xảy ra khi:

\(x - 20 = \dfrac{{400}}{{x - 20}} \Rightarrow x - 20 = 20 \Rightarrow x = 40cm\)