Đáp án:

 1) \(\left[ \begin{array}{l}
x = 5\\
x =  - 1
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
1)|{x^2} - 4x - 5| = {x^2} - 4x - 5\\
 \to \left[ \begin{array}{l}
{x^2} - 4x - 5 = {x^2} - 4x - 5\left( {ld} \right)\\
{x^2} - 4x - 5 =  - {x^2} + 4x + 5\left( {DK: - 1 \le x \le 5} \right)
\end{array} \right.\\
 \to 2{x^2} - 8x - 10 = 0\\
 \to \left[ \begin{array}{l}
x = 5\\
x =  - 1
\end{array} \right.\\
2){x^4} - 3{x^2} - 4 = 0\\
 \to \left[ \begin{array}{l}
{x^2} = 4\\
{x^2} =  - 1\left( l \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 2\\
x =  - 2
\end{array} \right.\\
3)|{x^2} - 4x - 5| = 0\\
 \to {x^2} - 4x - 5 = 0\\
 \to \left[ \begin{array}{l}
x = 5\\
x =  - 1
\end{array} \right.\\
4)\left( {x + 4} \right)\left( {x + 1} \right) - 3\sqrt {{x^2} + 5x - 2}  = 6\\
 \to \left( {{x^2} + 5x + 4} \right) - 3\sqrt {{x^2} + 5x - 2}  = 6\\
Đặt:\sqrt {{x^2} + 5x - 2}  = t\left( {t \ge 0} \right)\\
 \to {x^2} + 5x - 2 = {t^2}\\
 \to {x^2} + 5x = {t^2} + 2\\
Pt \to \left( {{t^2} + 6} \right) - 3t = 6\\
 \to t\left( {t - 3} \right) = 0\\
 \to \left[ \begin{array}{l}
t = 0\\
t = 3
\end{array} \right.\\
 \to \left[ \begin{array}{l}
{x^2} + 5x - 2 = 0\\
{x^2} + 5x - 2 = 9
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{ - 5 + \sqrt {33} }}{2}\\
x = \dfrac{{ - 5 - \sqrt {33} }}{2}\\
x = \dfrac{{ - 5 + \sqrt {69} }}{2}\\
x = \dfrac{{ - 5 - \sqrt {69} }}{2}
\end{array} \right.\\
5)|2x - 5| = |2{x^2} - 7x + 5|\\
 \to \left[ \begin{array}{l}
2x - 5 = 2{x^2} - 7x + 5\left( {DK:x \ge \dfrac{5}{2}} \right)\\
 - 2x + 5 = 2{x^2} - 7x + 5\left( {DK:x < \dfrac{5}{2}} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = 2\left( l \right)\\
x = 0
\end{array} \right.
\end{array}\)