Question

Solve: $(x^2-4)\sqrt{x^2-1}<0$

Answer 1

$(x^2-4)\sqrt{x^2-1}<0x^2-1>0x^2>1x\in (-1,-\infty )\bigsqcup (1,\infty )......\left(1\right)$

as $\sqrt{x^2-1}>0$

so $x^2-4<0x^2<4x\in (-2,2).....\left(2\right)$

combining ($1$) and ($2$)

we get $x\in (-2,-1)\bigsqcup (1,2)$