Giải thích các bước giải:
\(\begin{array}{l}
d\, \cap Ox = \left\{ A \right\} \Rightarrow A\left( {\dfrac{2}{{2m + 1}};0} \right)\\
d\, \cap Oy = \left\{ B \right\} \Rightarrow B\left( {0; - 2} \right)\\
\Rightarrow OA = \dfrac{2}{{\left| {2m + 1} \right|}};OB = 2\\
a)\,Ke\,OH \bot AB\,tai\,H\\
Khoang\,cach\,\,OH = \sqrt 2 \\
\dfrac{1}{{O{H^2}}} = \dfrac{1}{{O{A^2}}} + \dfrac{1}{{O{B^2}}}\\
\Leftrightarrow \dfrac{1}{2} = \dfrac{{{{\left( {2m + 1} \right)}^2}}}{4} + \dfrac{1}{4}\\
\Leftrightarrow {\left( {2m + 1} \right)^2} = 1\\
\Leftrightarrow \left[ \begin{array}{l}
2m + 1 = 1\\
2m + 1 = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
m = 0\\
m = - 1
\end{array} \right.\left( {tm} \right)\\
b)\,{S_{OAB}} = \dfrac{1}{2}OA.OB\\
\Rightarrow \dfrac{1}{2}.2.\dfrac{2}{{\left| {2m + 1} \right|}} = \dfrac{1}{2}\\
\Rightarrow \left| {2m + 1} \right| = 4\\
\Leftrightarrow \left[ \begin{array}{l}
2m + 1 = 4\\
2m + 1 = - 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
m = \dfrac{3}{2}\\
m = - \dfrac{5}{2}
\end{array} \right.
\end{array}\)
