Giải các bất phương trình :
$\begin{array}{l}
1)\,\,\,\,\,{\log _3}\left( {5{x^2} + 6x + 1} \right) \le 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\
2)\,\,\,\,{\log _{12}}\left( {6{x^2} - 48x + 54} \right) \le 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\
3)\,\,\,\,{\log _{21}}\left( {{x^2} + 2x - 3} \right) \le 1\,\,\,\\
4)\,\,\,\,{\log _2}\left( {{x^2} - 4x - 5} \right) \le 4
\end{array}$
$1)$
Điều kiện: $ 0 < 5{x^2} + 6x + 1 .$
Khi đó ta có 
$(1) \Leftrightarrow 0 < 5{x^2} + 6x + 1 \le 1\,\,\,\,\,\,$
$\Leftrightarrow \left\{ \begin{array}{l} (5x+1)(x+1)>0\\ x(5x+6)\leq 0\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} x<-1 \vee x>-\frac{1}{5}\\ -\frac{6}{5}\leq x\leq 0\end{array} \right.$
$\Leftrightarrow  \left[ \begin{array}{l}
 - \frac{6}{5} \le x <  - 1\\
 - \frac{1}{5} < x \le 0
\end{array} \right. $ (TMĐK)
Vậy BPT đã cho có nghiệm :$\left[ \begin{array}{l}
 - \frac{6}{5} \le x <  - 1\\
 - \frac{1}{5} < x \le 0
\end{array} \right.$

$2)\,\,$
Điều kiện: $6x^2-48x+54>0\Leftrightarrow (x-4+\sqrt7)(x-4-\sqrt7)>0$
                                                           $\Leftrightarrow \left[ \begin{array}{l} x>4+\sqrt7\\ x<4-\sqrt7 \end{array} \right.$
Khi đó BPT đã cho trở thành
$6x^2-48x+54\leq 12^2$
$\Leftrightarrow 6x^2-48x-90\leq 0$
$\Leftrightarrow x^2-8x-15\leq 0$
$\Leftrightarrow x^2-8x+16\leq 31$
$\Leftrightarrow (x-4)^2\leq 31$
$\Leftrightarrow -\sqrt{31}\leq x-4\leq \sqrt{31}$
$\Leftrightarrow 4-\sqrt{31}\leq x\leq 4+\sqrt{31}$
Kết hợp với điều kiện suy ra nghiệm của BPT là:
$$\left[ \begin{array}{l}
4 + \sqrt {7} < x \le 4 + \sqrt {31}  \\
4 - \sqrt {31}  \le x < 4 - \sqrt 7
\end{array} \right.$$
$3)$
Điều kiện: $x^2+2x-3>0\Leftrightarrow (x+3)(x-1)>0\Leftrightarrow \left[\begin{array}{l} x>1\\ x<-3 \end{array} \right.$
Khi đó BPT đã cho trở thành
$x^2+2x-3\leq 21$
$\Leftrightarrow x^2+2x-24\leq 0$
$\Leftrightarrow (x+6)(x-4)\leq 0\Leftrightarrow -6\leq x\leq 4.$
Kết hợp với điều kiện vậy nghiệm BPT đã cho là
$$ - 6 \le x <  - 3\,\,\,;\,\,\,1 < x \le 4$$
$4)$
Điều kiện: $x^2-4x-5>0\Leftrightarrow (x-5)(x+1)>0\Leftrightarrow \left[ \begin{array}{l} x>5\\ x<-1 \end{array} \right.$
Khi đó BPT đã cho trở thành
$x^2-4x-5\leq 2^4$
$\Leftrightarrow x^2-4x-21\leq 0$
$\Leftrightarrow (x-7)(x+3)\leq 0$
$\Leftrightarrow 3\leq x\leq 7$
Kết hợp với điều kiện vậy nghiệm BPT đã cho là 
$$ - 3 \le x < 1\,\,\,\,;\,\,5 < x \le 7$$

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