Đặt \(3^x=t>0\)
\(\Rightarrow t+3=m\sqrt{t^2+1}\Rightarrow m=\dfrac{t+3}{\sqrt{t^2+1}}\)
Xét hàm \(f\left(t\right)=\dfrac{t+3}{\sqrt{t^2+1}}\) với \(t>0\)
\(f'\left(t\right)=\dfrac{\sqrt{t^2+1}-\dfrac{\left(t+3\right).t}{\sqrt{t^2+1}}}{t^2+1}=\dfrac{1-3t}{\left(t^2+1\right)\sqrt{t^2+1}}\)
\(f'\left(t\right)=0\Rightarrow t=\dfrac{1}{3}\)
\(f\left(0\right)=3\) ; \(f\left(\dfrac{1}{3}\right)=\sqrt{10}\); \(\lim\limits_{t\rightarrow+\infty}f\left(t\right)=1\)
BBT:
Từ BBT ta thấy pt có đúng 1 nghiệm khi: \(\left[{}\begin{matrix}m=\sqrt{10}\\1< m\le3\end{matrix}\right.\)