ĐKXĐ: \(-2\le x\le2\)
\(\Leftrightarrow\dfrac{2x+4-4\left(2-x\right)}{\sqrt{2x+4}+2\sqrt{2-x}}-\dfrac{6x-4}{5\sqrt{x^2+1}}\ge0\)
\(\Leftrightarrow\dfrac{6x-4}{\sqrt{2x+4}+2\sqrt{2-x}}-\dfrac{6x-4}{5\sqrt{x^2+1}}\ge0\)
\(\Leftrightarrow\left(6x-4\right)\left(\dfrac{1}{\sqrt{2x+4}+2\sqrt{2-x}}-\dfrac{1}{5\sqrt{x^2+1}}\right)\ge0\) (1)
Ta có:
\(5\sqrt{x^2+1}\ge5.\sqrt{0+1}=5=\sqrt{25}\)
\(\sqrt{2x+4}+\sqrt{2}.\sqrt{4-2x}\le\sqrt{\left(1+2\right)\left(2x+4+4-2x\right)}=\sqrt{24}\)
\(\Rightarrow5\sqrt{x^2+1}>\sqrt{2x+4}+2\sqrt{2-x}\)
\(\Rightarrow\dfrac{1}{\sqrt{2x+4}+2\sqrt{2-x}}-\dfrac{1}{5\sqrt{x^2+1}}>0;\forall x\)
Nên (1) tương đương:
\(6x-4\ge0\Rightarrow x\ge\dfrac{2}{3}\)
\(\Rightarrow\dfrac{2}{3}\le x\le2\)
