Tọa độ điểm A là:
\(\left\{{}\begin{matrix}y=0\\mx=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=\dfrac{5}{m}\end{matrix}\right.\)
Tọa độ điểm B là:
\(\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{m+1}\end{matrix}\right.\)
Theo đề, ta có: \(\dfrac{1}{\left|\dfrac{5}{m}\right|^2}+\dfrac{1}{\left|\dfrac{5}{m+1}\right|^2}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{\dfrac{25}{\left|m^2\right|}}+\dfrac{1}{\dfrac{25}{\left(m+1\right)^2}}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{m^2}{25}+\dfrac{\left(m+1\right)^2}{25}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{2m^2+2m+1}{25}=\dfrac{1}{8}\)
\(\Leftrightarrow16m^2+16m+8-25=0\)
\(\Leftrightarrow16m^2+16m-17=0\)
hay \(m\in\left\{\dfrac{-2+\sqrt{21}}{4};\dfrac{-2-\sqrt{21}}{4}\right\}\)
