Đk: $x\neq 1$Bpt $\Leftrightarrow x+2\geq \frac{\sqrt{2(x^4-x^2+1)}-1}{x-1}$ $(*)$
TH1: Nếu $x>1$ thì
$(*)\Leftrightarrow (x+2)(x-1)\geq \sqrt{2(x^4-x^2+1)}-1$
$\Leftrightarrow x^2+x-1\geq \sqrt{2(x^4-x^2+1)}$
$\Leftrightarrow \begin{cases}x^2+x-1\geq 0\\ (x^2+x-1)^2\geq 2(x^4-x^2+1)\end{cases}$
$\Leftrightarrow \begin{cases}x^2+x-1\geq 0\\ x^4-2x^3-x^2+2x+1\leq 0\end{cases}$
$\Leftrightarrow \begin{cases}x^2+x-1\geq 0\\ (x^2-x-1)^2\leq 0\end{cases}\Leftrightarrow \begin{cases}x^2+x-1\geq 0\\ x^2-x-1=0 \end{cases}$
$\Leftrightarrow x=\frac{1+\sqrt{5}}{2}$
TH2: Nếu $x<1$ thì $\sqrt{2(x^4-x^2+1)}-1>0$
$(*)\Leftrightarrow (x+2)(x-1)\leq \sqrt{2(x^4-x^2+1)}-1$
$\Leftrightarrow \begin{cases} -2\leq x<1\\ \begin{cases}x<-2 \\ (x^2-x-1)^2\geq 0\end{cases} \end{cases}$
$\Leftrightarrow x<1$
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