Đặt \(\sqrt{\left(x+4\right)\left(2-x\right)}=t\Rightarrow-x^2-2x+8=t^2\)
\(\Rightarrow x^2+2x=8-t^2\)
Ta có: \(\sqrt{\left(x+4\right)\left(2-x\right)}\le\dfrac{1}{2}\left(x+4+2-x\right)=3\Rightarrow0\le t\le3\)
BPT trở thành:
\(t\ge8-t^2+2m-1\Leftrightarrow t^2+t-7\ge2m\) ; \(\forall t\in\left[0;3\right]\)
\(\Leftrightarrow2m\le\min\limits_{\left[0;3\right]}\left(t^2+t-7\right)\)
Xét hàm \(f\left(t\right)=t^2+t-7\) trên \(\left[0;3\right]\)
\(-\dfrac{b}{2a}=-\dfrac{1}{2}< 0\Rightarrow f\left(t\right)\) đồng biến trên \(\left[0;3\right]\)
\(\Rightarrow\min\limits_{\left[0;3\right]}f\left(t\right)=f\left(0\right)=-7\)
\(\Rightarrow2m\le-7\Rightarrow m\le-\dfrac{7}{2}\)
