ĐKXĐ: \(-\frac{1}{2}\le x\le3\)
\(\Leftrightarrow-2x^2+5x+3+\sqrt{-2x^2+5x+3}-3=m\)
Đặt \(\sqrt{-2x^2+5x+3}=t\Rightarrow0\le t\le\frac{7\sqrt{2}}{4}\)
\(\Rightarrow t^2+t-3=m\)
Xét \(f\left(t\right)=t^2+t-3\) trên \(\left[0;\frac{7\sqrt{2}}{4}\right]\)
\(-\frac{b}{2a}=-\frac{1}{2}< 0\Rightarrow f\left(t\right)\) đồng biến trên \(\left[0;\frac{7\sqrt{2}}{4}\right]\)
\(\Rightarrow f\left(0\right)\le f\left(t\right)\le f\left(\frac{7\sqrt{2}}{4}\right)\Leftrightarrow-3\le f\left(t\right)\le\frac{25+14\sqrt{2}}{8}\)
\(\Rightarrow-3\le m\le\frac{25+14\sqrt{2}}{8}\)