TH1 : \(x\ge m\)
\(PT\Leftrightarrow2x^2+2\left(m+1\right)x-m^2-1=x^2-2mx+m^2\)
\(\Leftrightarrow x^2+2\left(2m+1\right)x-2m^2-1=0\)
Có \(\Delta^,=b^{,2}-ac=4m^2+4m+1+2m^2+1=6m^2+4m+2\)
- Thấy \(\Delta^,\ge\dfrac{4}{3}>0\)
- Nên để PT có nghiệm thì \(x_1>x_2>m\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(m\right)>0\\-\left(2m+1\right)>m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2+2\left(2m+1\right)m-2m^2-1>0\\-\left(2m+1\right)-m>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3m^2+2m-1>0\\3m+1< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< -1\\m>\dfrac{1}{3}\end{matrix}\right.\\m< -\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow m< -1\)
TH2 : \(\left\{{}\begin{matrix}x< m\\2x^2+2\left(m+1\right)x-m^2-1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< m\\\Delta^,=3m^2+2m+3\le0\end{matrix}\right.\)
<=> Loại .
Vậy để .... <=> m < - 1