Đáp án:

$\begin{array}{l}
1)a)\left| {4x + 5} \right| = \left| {x - 3} \right|\\
 \Rightarrow \left[ \begin{array}{l}
4x + 5 = x - 3\\
4x + 5 = 3 - x
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
4x - x =  - 3 - 5\\
4x + x = 3 - 5
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
3x =  - 8\\
5x =  - 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \frac{{ - 8}}{3}\\
x =  - \frac{2}{5}
\end{array} \right.\\
\text{Vậy}\,x =  - \frac{8}{3};x = \frac{{ - 2}}{5}\\
b)Dkxd:2x - 3 \ge 0 \Rightarrow x \ge \frac{3}{2}\\
\sqrt {2x - 3}  = 3\\
 \Rightarrow 2x - 3 = 9\\
 \Rightarrow 2x = 9 + 3\\
 \Rightarrow 2x = 12\\
 \Rightarrow x = 12:2\\
 \Rightarrow x = 6\left( {tmdk} \right)\\
\text{Vậy}\,x = 6\\
2)a)m = 0\\
 \Rightarrow {x^2} - 2x + 3 - \sqrt {{x^2} - 2x + 5}  = 0\\
\text{Đặt}:\sqrt {{x^2} - 2x + 5}  = a\left( {a \ge 0} \right)\\
 \Rightarrow {x^2} - 2x + 5 = {a^2}\\
 \Rightarrow {x^2} - 2x + 3 = {a^2} - 2\\
 \Rightarrow {a^2} - 2 - a = 0\\
 \Rightarrow {a^2} - a - 2 = 0\\
 \Rightarrow \left( {a - 2} \right)\left( {a + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
a = 2\left( {tm} \right)\\
a =  - 1\left( {ktm} \right)
\end{array} \right.\\
 \Rightarrow a = 2\\
 \Rightarrow \sqrt {{x^2} - 2x + 5}  = 2\\
 \Rightarrow {x^2} - 2x + 5 = 4\\
 \Rightarrow {x^2} - 2x + 1 = 0\\
 \Rightarrow {\left( {x - 1} \right)^2} = 0\\
 \Rightarrow x = 1\\
\text{Vậy}\,x = 1\\
3)b)D\left( {x;y} \right)\\
\overrightarrow {AB}  = \left( {3;1} \right);\overrightarrow {DC}  = \left( { - 1 - x;4 - y} \right)\\
\text{ABCD là hình bình hành nên:}\\
 \Rightarrow \overrightarrow {AB}  = \overrightarrow {DC} \\
 \Rightarrow \left( {3;1} \right) = \left( { - 1 - x;4 - y} \right)\\
 \Rightarrow \left\{ \begin{array}{l}
3 =  - 1 - x\\
1 = 4 - y
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x =  - 4\\
y = 3
\end{array} \right.\\
 \Rightarrow D\left( { - 4;3} \right)
\end{array}$