Giải thích các bước giải:
Từ đề ta suy ra :
$\log_2(2-x)+\log_2(x+1)+1\le 2\log_2(m-\dfrac x2+4(\sqrt{2-x}+\sqrt{2x+2}))$
$\to \log_2(2-x)2(x+1)\le 2\log_2(m-\dfrac x2+4(\sqrt{2-x}+\sqrt{2x+2}))$
$\to \log_2(2-x)(2x+2)\le 2\log_2(m-\dfrac x2+4(\sqrt{2-x}+\sqrt{2x+2}))$
$\to \dfrac 12\log_2(2-x)(2x+2)\le \log_2(m-\dfrac x2+4(\sqrt{2-x}+\sqrt{2x+2}))$
$\to \log_2\sqrt{2-x}.\sqrt{2x+2}\le \log_2(m-\dfrac x2+4(\sqrt{2-x}+\sqrt{2x+2}))$
$\to \sqrt{2-x}.\sqrt{2x+2}\le m-\dfrac x2+4(\sqrt{2-x}+\sqrt{2x+2})$
$\to \dfrac x2-4(\sqrt{2-x}+\sqrt{2x+2})+\sqrt{2-x}.\sqrt{2x+2}\le m$
$\to \dfrac{x}{2}-4(\sqrt{2-x}+\sqrt{2x+2})+\sqrt{2-x}.\sqrt{2x+2}\le m$
Đặt $f(x)=\dfrac{x}{2}-4(\sqrt{2-x}+\sqrt{2x+2})+\sqrt{2-x}.\sqrt{2x+2}$
$\to f'(x)=\dfrac{1}{2}-4\left(-\dfrac{1}{2\sqrt{2-x}}+\dfrac{1}{\sqrt{2x+2}}\right)-\dfrac{\sqrt{2x+2}}{2\sqrt{2-x}}+\sqrt{\dfrac{2-x}{2x+2}}$
$\to f'(x)=0\to x=1\to x=1$ là min của f(x) (khảo sát bằng máy tính )
$\to f(x)\ge -9.5$
$\to m\ge -9.5\to m=-9.5\to C$
