ĐK: \(\hept{\begin{cases}1-\frac{2}{x}\ge0\\2x-\frac{8}{x}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{x-2}{x}\ge0\\\frac{2x^2-8}{x}\ge0\end{cases}}\)
<=> \(-2\le x< 0\) hoặc \(x\ge2\)
TH1: \(-2\le x< 0\)
Bất phương trình đúng
TH2: \(x\ge2\)(@@)
bất pt <=> \(2\sqrt{\frac{x-2}{x}}+\sqrt{\frac{2\left(x-2\right)\left(x+2\right)}{x}}\ge x\)
<=> \(\sqrt{\frac{x-2}{x}}\left(2+\sqrt{2\left(x+2\right)}\right)\ge x\)
<=> \(\sqrt{\frac{x-2}{x}}\left(\frac{2x}{\sqrt{2\left(x+2\right)}-2}\right)\ge x\)
<=> \(2\sqrt{\frac{x-2}{x}}+2\ge\sqrt{2\left(x+2\right)}\)
<=> \(4\left(1-\frac{2}{x}\right)+4+8\sqrt{1-\frac{2}{x}}\ge2x+4\)
<=> \(4\sqrt{1-\frac{2}{x}}\ge x-2+\frac{4}{x}\)
<=> \(16\left(1-\frac{2}{x}\right)\ge x^2+4+\frac{16}{x^2}-4x+8-\frac{16}{x}\)
<=> \(4\ge x^2+\frac{16}{x^2}-4x+\frac{16}{x}\)
<=> \(\left(x-\frac{4}{x}\right)^2-4\left(x-\frac{4}{x}\right)+4\le0\)
<=> \(\left(x-\frac{4}{x}+2\right)^2\le0\) vô nghiệm vì x > 2 => \(x-\frac{4}{x}+2>2\)
Vậy -2 \(\le\) x < 0
