a, ta có \(\dfrac{1}{f}=\dfrac{1}{d}+\dfrac{1}{d'}\)
Vị trí ban đầu \(\dfrac{1}{f}=\dfrac{d_1+d'_1}{d_1d'_1}\left(1\right)\)
Vị trị sau \(\dfrac{1}{f}=\dfrac{d_2+d'_2}{d_2d'_2}\left(2\right)\)
Theo gt, ta đc
\(\dfrac{A_1B_1}{A_2B_2}=\dfrac{1}{10}\Rightarrow\dfrac{A_1B_1}{AB}.\dfrac{AB}{A_2B_2}=\dfrac{1}{20}\\ \Leftrightarrow\dfrac{d'_1}{d_1}=\dfrac{d_2}{d'_2}=\dfrac{1}{10}\Leftrightarrow10d_1d_2=d_1d'_2\)
Theo gt
\(10d_1'=\left(d_1-45\right)=d_1\left(d_1'+18\right)\\ \Leftrightarrow9d_1d'_1-459d'_1-18d_1=0\\ \Leftrightarrow d_1=\dfrac{450d'_1}{9d_1'-18}\left(3\right)\)
Từ (1) và (2)
\(\dfrac{d_1+d'_1}{d_1d'_1}=\dfrac{d_2d'_2}{d_2d'_2}\\ \Leftrightarrow d_2d'_2\left(d_1+d'_1\right)=d_1d'_1\left(d_2+d'_2\right)\\ \Leftrightarrow d_2d'_2d_1+d_2d'_2d'_1=d_1d'_1d_2+d_1d'_1d_2'\\ \Leftrightarrow d'_1d_2\left(d'_2-d_1\right)=d_1d'_2\left(d'_1-d_2\right)\\ \Leftrightarrow d'_1d_2\left(d'_2-d_1\right)=10d'_1d_2\left(d'_1-d_2\right)\\ \Leftrightarrow d_2'-d_1=10d'_1-10d_2\\ \Leftrightarrow d'_1+18-d_1=10d'_1-10d_1+450\\ \Leftrightarrow d_1=\dfrac{9d'_1+432}{9}\left(4\right)\)
Từ (3) và (4)
\(\Leftrightarrow\dfrac{450d'_1}{9d'_1-18}=\dfrac{9d'_1+432}{9}\\ \Rightarrow d'_1=12cm\\ d_1=60cm\\ f=\dfrac{d_1d'_1}{d_1+d'_1}=10cm\)
