2 câu trả lời
$\dfrac{1}{d}+\dfrac{1}{d'}=\dfrac{1}{f}$
$\to\begin{cases} d=\dfrac{f.d'}{d'-f}\\d'=\dfrac{fd}{d-f}\end{cases}$
Mà $k=\dfrac{-d'}{d}$
$\Rightarrow d'=-kd$
$\Rightarrow d=\dfrac{f-kd}{-kd-f}$
$\Rightarrow kd+f=fk$
$\Longrightarrow \begin{cases} d=f-\dfrac{f}{k}\\d'=f-fk\end{cases}$
Đáp án:
Ta có:
\(\begin{array}{l}
\dfrac{1}{f} = \dfrac{1}{d} + \dfrac{1}{{d'}} \Rightarrow dd' = d'f + df \Rightarrow d' = \dfrac{{df}}{{d - f}}\\
K = - \dfrac{{d'}}{d} = - \dfrac{f}{{d - f}} = \dfrac{f}{{f - d}}
\end{array}\)
