$\{\begin{array}{c}\frac{x^4-6x^2+9}{x-1}\le 0\\ \frac{\mid x-8\mid -1}{\sqrt{x+2}}>0\end{array}$

• $\frac{x^4-6x^2+9}{x-1}\le 0\Rightarrow \frac{x^4-6x^2+9}{1-x}\ge 0$
  • $x^4-6x^2+9\ge 0\Rightarrow {(x^2-3)}^2\ge 0\Rightarrow \forall x\in ℝ$
  • $1-x>0\Rightarrow x<1$

  $x<1$

• $\frac{\mid x-8\mid -1}{\sqrt{x+2}}>0$
  • $\mid x-8\mid -1>0\Rightarrow \{\begin{array}{c}x\ge 8\\ x-8-1>0\end{array}\vee \{\begin{array}{c}x<8\\ 8-x-1>0\end{array}\Rightarrow \{\begin{array}{c}x\ge 8\\ x>9\end{array}\vee \{\begin{array}{c}x<8\\ x<7\end{array}\Rightarrow x<7\vee x>9$
  • $\sqrt{x+2}>0\Rightarrow x>-2$

  $-2<x<7\vee x>9$

$-2<x<1$