Giải các bất phương trình :
$1)\,\,\,{\log _x}\left( {1 + {a^2}} \right) < 0\,\,\,\,\,\,\,\,\,\,\,\,(1)$ ( $a$ là tham số)
$2)\,\,\frac{1}{{{{\log }_a}x}} > 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$
$1)$      Điều kiện: $1\ne x>0$
xét $a = 0$ hoặc $a \ne 0$
   $a = 0$ : $(1)$ vô nghiệm
   $a \ne 0$ : $1 + {a^2} > 1$
Suy ra $\log_x(1+a^2)<0\Leftrightarrow x<1 $
                        ĐS : $0 < x < 1$

$2)$Điều kiện: $\,\,\,a > 0,\,\,a \ne 1,\,\,\,x > 0,\,\,x \ne 1$
     Xét $2$ trường hợp:
$a > 1        :$   Ta có:   $\frac1{\log_a x} >1\Leftrightarrow 0<\log_a x<1\Leftrightarrow 1  < x <a$
$0 < a < 1:$   Ta có : $\frac1{\log_a x} >1\Leftrightarrow 0<\log_a x<1\Leftrightarrow a < x < 1$
Kết luận:
Với $a > 1        :$  nghiệm của bất phương trình là: $1  < x <a$
 Với $0 < a < 1:$  nghiệm của bất phương trình là: $a < x < 1$
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