a/
\(\Leftrightarrow\sqrt{x^2-1}=x+m\) (\(x\ge-m\))
\(\Leftrightarrow x^2-1=x^2+2mx+m^2\)
\(\Leftrightarrow2mx=-1-m^2\) (\(m=0\) pt vô nghiệm)
\(\Rightarrow x=\frac{-1-m^2}{2m}\)
\(\Rightarrow\frac{-1-m^2}{2m}\ge-m\Leftrightarrow\frac{1+m^2}{2m}-m\le0\)
\(\Leftrightarrow\frac{1-m^2}{2m}\le0\)
\(\Rightarrow\left[{}\begin{matrix}-1\le m< 0\\m\ge1\end{matrix}\right.\)
b/ \(x\ge m\)
\(\Leftrightarrow2x^2+mx-3=\left(x-m\right)^2\)
\(\Leftrightarrow2x^2+mx-3=x^2-2mx+m^2\)
\(\Leftrightarrow x^2+3mx-m^2-3=0\) (1)
\(ac< 0\Rightarrow\left(1\right)\) luôn luôn có nghiệm
Để (1) có 2 nghiệm thỏa mãn \(x_1< x_2< m\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(m\right)>0\\\frac{S}{2}< m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2+3m^2-m^2-3>0\\-\frac{3m}{2}< m\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m^2>1\\m>0\end{matrix}\right.\) \(\Rightarrow m>1\)
Vậy để pt đã cho có nghiệm thì \(m\le1\)
