Ta có: \(f\left(x\right)=\sqrt{7-2x}+\sqrt{3x+4}\)
Điều kiện: \(-\dfrac{4}{3} \leq x \le \dfrac{7}{2}\)
\({y^2} = {\left( {\sqrt {7 - 2x} + \sqrt {3x + 4} } \right)^2} = x + 11 + 2\sqrt {\left( {7 - 2x} \right)\left( {3x + 4} \right)} = \dfrac{1}{3}\left( {3x + 4} \right) + 2\sqrt {\left( {7 - 2x} \right)\left( {3x + 4} \right)} + \dfrac{{29}}{3}\)
Vì: \(\left\{{}\begin{matrix}3x+4\ge0\\\sqrt{\left(7-2x\right)\left(3x+4\right)}\ge0\end{matrix}\right.\) \(\forall x \in \left[ { - \dfrac{4}{3};\dfrac{7}{2}} \right] \Rightarrow {f^2}\left( x \right) \ge \dfrac{{29}}{3} \Rightarrow f\left( x \right) \ge \dfrac{{\sqrt {87} }}{3}\)
Dấu \("="\) xảy ra \(\Leftrightarrow x=-\dfrac{4}{3}\). Vậy \(m=\dfrac{\sqrt{87}}{3}\)