Đáp án:

$\begin{array}{l}
1)2{x^2} - 3x - 5 = 0\\
 \Rightarrow 2{x^2} - 5x + 2x - 5 = 0\\
 \Rightarrow \left( {2x - 5} \right)\left( {x + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x =  - 1
\end{array} \right.\\
b)x + 4\sqrt x  - 5 = 0\\
 \Rightarrow x + 5\sqrt x  - \sqrt x  - 5 = 0\\
 \Rightarrow \left( {\sqrt x  + 5} \right)\left( {\sqrt x  - 1} \right) = 0\\
 \Rightarrow \sqrt x  = 1\\
 \Rightarrow x = 1\\
c)\left\{ \begin{array}{l}
2x - y = 3\\
3x + 2y = 8
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
4x - 2y = 6\\
3x + 2y = 8
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
7x = 14\\
y = 2x - 3
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = 1
\end{array} \right.\\
2)a)Dkxd:\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
A = \left( {\dfrac{1}{{\sqrt x  + 1}} - \dfrac{1}{{x + \sqrt x }}} \right):\dfrac{{\sqrt x  - 1}}{{x + 2\sqrt x  + 1}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}.\dfrac{{{{\left( {\sqrt x  + 1} \right)}^2}}}{{\sqrt x  - 1}}\\
 = \dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
 = 2\\
 \Rightarrow \dfrac{{\sqrt x  + 1}}{{\sqrt x }} = 2\\
 \Rightarrow 2\sqrt x  = \sqrt x  + 1\\
 \Rightarrow \sqrt x  = 1\\
 \Rightarrow x = 1\left( {ktm} \right)\\
c)A \le \dfrac{3}{4}\\
 \Rightarrow \dfrac{{\sqrt x  + 1}}{{\sqrt x }} \le \dfrac{3}{4}\\
 \Rightarrow \dfrac{{4\sqrt x  + 4 - 3\sqrt x }}{{4\sqrt x }} \le 0\\
 \Rightarrow \dfrac{{\sqrt x  + 4}}{{4\sqrt x }} \le 0\left( {vo\,nghiem} \right)\\
 \Rightarrow ko\,co\,x\\
3)a)\Delta  = {\left( {2m - 3} \right)^2} - 4.\left( {2m - 4} \right)\\
 = 4{m^2} - 12m + 9 - 8m + 16\\
 = 4{m^2} - 20m + 25\\
 = 4.\left( {{m^2} - 5m + \dfrac{{25}}{4}} \right)\\
 = 4.{\left( {m - \dfrac{5}{2}} \right)^2} \ge 0
\end{array}$

Vậy pt có 2 nghiệm với mọi m.

b)

$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m - 3\\
{x_1}{x_2} = 2m - 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x_1^2 - \left( {2m - 3} \right){x_1} + 2m - 4 = 0\\
x_2^2 - \left( {2m - 3} \right){x_2} + 2m - 4 = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x_1^2 - 2m{x_1} + 2m - 3 =  - 3{x_1} + 1\\
x_2^2 - 2m{x_2} + x + 2m - 3 =  - 3{x_2} + 1
\end{array} \right.\\
 \Rightarrow \left( { - 3{x_1} + 1 + {x_2}} \right)\left( { - 3{x_2} + 1 + {x_1}} \right) = 30\\
 \Rightarrow 9{x_1}{x_2} - 3x_1^2 - 3{x_1} - 3{x_2} + 1 + {x_1} - 3x_2^2 + {x_2} + {x_1}{x_2} = 30\\
 \Rightarrow  - 3\left( {x_1^2 + x_2^2} \right) + 10{x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) - 29 = 0\\
 \Rightarrow  - 3.{\left( {{x_1} + {x_2}} \right)^2} + 6{x_1}{x_2} + 10{x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) - 29 = 0\\
 \Rightarrow  - 3.{\left( {2m - 3} \right)^2} + 16.\left( {2m - 4} \right) - 2.\left( {2m - 3} \right) - 29 = 0\\
 \Rightarrow  - 12{m^2} + 36m - 27 + 32m - 64 - 4m + 6 - 29 = 0\\
 \Rightarrow 12{m^2} - 64m + 114 = 0\left( {vô\,nghiệm} \right)
\end{array}$

Vậy ko có giá trị của m thỏa mãn đề bài.