Đặt \(\sqrt{log_3^2x+1}=t\in\left[1;\sqrt{2}\right]\)
\(\Rightarrow f\left(t\right)=t^2+t-1=3m\)
\(f'\left(t\right)=2t+1>o;\forall t\in\left[1;\sqrt{2}\right]\)
\(f\left(1\right)=1;f\left(\sqrt{2}\right)=1+\sqrt{2}\)
\(\Rightarrow1\le3m\le1+\sqrt{2}\)