giải bpt:
$\dfrac{2+4x}{|x^2-1|-2}$<1
thank.....
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TH1:
$\left\{\begin{matrix}x^2-1<0\\\dfrac{x+4x}{1-x^2-2}-1<0\end{matrix}\right.$
\Leftrightarrow
$\left\{\begin{matrix}-1<x<1\\\dfrac{x^2+4x+3}{-x^2-1}<0\end{matrix}\right.$
\Leftrightarrow
$\left\{\begin{matrix}-1<x<1\\x<-3 v x>-1\end{matrix}\right.$
\Leftrightarrow -1<x<1
TH2:
$\left\{\begin{matrix}x^2-1\geq0\\\dfrac{x+4x}{x^2-1-2}-1<0\end{matrix}\right.$
\Leftrightarrow $\left\{\begin{matrix}x\leq-1 v x\geq1\\\dfrac{x^2-4x-5}{x^2-3}<0\end{matrix}\right.$
\Leftrightarrow
$\left\{\begin{matrix}x\leq-1 v x\geq1\\x<-\sqrt{3} v -1<x<\sqrt{3} v x>5\end{matrix}\right.$
\Leftrightarrow x<-$\sqrt{3}$ v -1\leqx<$\sqrt{3}$ v x>5
vậy từ 2 TH ta đc:
x<-$\sqrt{3}$ v -1\leqx<$\sqrt{3}$ v x>5