PT giao Ox và Oy:
\(y=0\Leftrightarrow x=\dfrac{2m+1}{m+3}\Leftrightarrow A\left(\dfrac{2m+1}{m+3};0\right)\Leftrightarrow OA=\left|\dfrac{2m+1}{m+3}\right|\\ x=0\Leftrightarrow y=-2m-1\Leftrightarrow B\left(0;-2m-1\right)\Leftrightarrow OB=\left|2m+1\right|\)
Gọi H là chân đường cao từ O tới (d)
Đặt \(OH^2=t\)
Áp dụng HTL: \(\dfrac{1}{OH^2}=\dfrac{1}{t}=\dfrac{1}{OA^2}+\dfrac{1}{OB^2}=\dfrac{\left(m+3\right)^2}{\left(2m+1\right)^2}+\dfrac{1}{\left(2m+1\right)^2}\)
\(\Leftrightarrow\dfrac{1}{t}=\dfrac{m^2+6m+10}{4m^2+4m+1}\\ \Leftrightarrow tm^2+6mt+10t=4m^2+4m+1\\ \Leftrightarrow m^2\left(t-4\right)+2m\left(3t-2\right)+10t-1=0\)
Vì PT bậc 2 ẩn m này có nghiệm nên \(\Delta'\ge0\)
\(\Leftrightarrow\left(3t-2\right)^2-\left(10t-1\right)\left(t-4\right)\ge0\\ \Leftrightarrow9t^2-12t+4-10t^2+41t-4\ge0\\ \Leftrightarrow-t^2+29t\ge0\\ \Leftrightarrow0\le t\le29\)
Do đó \(0\le OH\le\sqrt{29}\)
Dấu \("="\Leftrightarrow\dfrac{4m^2+4m+1}{m^2+6m+10}=29\Leftrightarrow25m^2+170m+289=0\)
\(\Leftrightarrow\left(5m+17\right)^2=0\Leftrightarrow m=-\dfrac{17}{5}\)
Vậy \(OH_{max}=\sqrt{29}\Leftrightarrow m=-\dfrac{17}{5}\)
