ĐKXĐ: \(x\ge3\)
\(\Leftrightarrow mx-\sqrt{x-3}=m+1\Leftrightarrow m\left(x-1\right)=\sqrt{x-3}+1\)
\(\Leftrightarrow m=\frac{\sqrt{x-3}+1}{x-1}\)
Đặt \(\sqrt{x-3}=t\ge0\) \(\Rightarrow x=t^2+3\Rightarrow m=\frac{t+1}{t^2+2}\)
Xét hàm \(f\left(t\right)=\frac{t+1}{t^2+2}\Rightarrow f'\left(t\right)=\frac{t^2+2-2t\left(t+1\right)}{\left(t^2+2\right)^2}=\frac{-t^2-2t+2}{\left(t^2+2\right)^2}\)
\(f'\left(t\right)=0\Rightarrow t=\sqrt{3}-1\)
Ta có \(f\left(\sqrt{3}-1\right)=\frac{1+\sqrt{3}}{4}\); \(\lim\limits_{t\rightarrow+\infty}\frac{t+1}{t^2+1}=0\); \(f\left(0\right)=\frac{1}{2}\)
Dựa vào BBT, để pt đã cho có 2 nghiệm pb thì \(\frac{1}{2}\le m< \frac{1+\sqrt{3}}{4}\)
