Đáp án:

 1) $S = \left\{ {\dfrac{{37}}{{43}}} \right\}$

2) $S = \left\{ {\dfrac{3}{2};4} \right\}$

3) $S = \left\{ { - 2;\dfrac{5}{2}} \right\}$

4) $S = \left\{ {1;6} \right\}$

Giải thích các bước giải:

$\begin{array}{l}
1)\dfrac{{x + 5}}{4} - \dfrac{{2x - 5}}{3} = \dfrac{{6x - 1}}{2} + \dfrac{{2x - 3}}{{12}}\\
 \Leftrightarrow \dfrac{{3\left( {x + 5} \right) - 4\left( {2x - 5} \right) - 6\left( {6x - 1} \right) - \left( {2x + 3} \right)}}{{12}} = 0\\
 \Leftrightarrow  - 43x + 37 = 0\\
 \Leftrightarrow x = \dfrac{{37}}{{43}}
\end{array}$

Vậy phương trình có tập nghiệm là: $S = \left\{ {\dfrac{{37}}{{43}}} \right\}$

$\begin{array}{l}
2)\left( {2x - 3} \right)\left( {x + 7} \right) = 4{x^2} - 9\\
 \Leftrightarrow \left( {2x - 3} \right)\left( {x + 7} \right) - \left( {2x - 3} \right)\left( {2x + 3} \right) = 0\\
 \Leftrightarrow \left( {2x - 3} \right)\left( { - x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 0\\
 - x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = 4
\end{array} \right.
\end{array}$

Vậy phương trình có tập nghiệm là: $S = \left\{ {\dfrac{3}{2};4} \right\}$

$\begin{array}{l}
3)2{x^2} - x - 10 = 0\\
 \Leftrightarrow \left( {x + 2} \right)\left( {2x - 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
2x - 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 2\\
x = \dfrac{5}{2}
\end{array} \right.
\end{array}$

Vậy phương trình có tập nghiệm là: $S = \left\{ { - 2;\dfrac{5}{2}} \right\}$

4) ĐKXĐ: $x\ne \{-3;4\}$

$\begin{array}{l}
\dfrac{{x + 4}}{{x + 3}} + \dfrac{{12x + 7}}{{\left( {x + 3} \right)\left( {x - 4} \right)}} = \dfrac{{1 - 2x}}{{4 - x}}\\
 \Leftrightarrow \dfrac{{\left( {x + 4} \right)\left( {x - 4} \right) + 12x + 7 - \left( {2x - 1} \right)\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x - 4} \right)}} = 0\\
 \Leftrightarrow  - {x^2} + 7x - 6 = 0\\
 \Leftrightarrow {x^2} - 7x + 6 = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x - 6} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x - 6 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1(tm)\\
x = 6(tm)
\end{array} \right.
\end{array}$

Vậy phương trình có tập nghiệm là: $S = \left\{ {1;6} \right\}$

`1,\frac{x+5}{4}-\frac{2x-5}{3}=\frac{6x-1}{2}+\frac{2x-3}{12}`

`⇔\frac{3(x+5)-4(2x-5)}{12}=\frac{6(6x-1)+2x-3}{12}`

`⇒3x+15-8x+20=36x-6+2x+3`

`⇔-43x=-38`

`⇔x=38/43`

`2,(2x-3)(x+7)=4x^2-9`

`⇔(2x-3)(x+7)-(2x-3)(2x+3)=0`

`⇔(2x-3)(x+7-2x-3)=0`

`⇔(2x-3)(4-x)=0`

`⇔` \(\left[ \begin{array}{l}2x-3=0\\4-x=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=4\end{array} \right.\) 

`3,2x^2-x-10=0`

`⇔2x^2-5x+4x-10=0`

`⇔2x(x+2)-5(x+2)=0`

`⇔(2x-5)(x+2)=0`

`⇔` \(\left[ \begin{array}{l}2x-5=0\\x+2=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=-2\end{array} \right.\) 

`4,\frac{x+4}{x+3}+\frac{12x+7}{(x+3)(x-4)}=\frac{1-2x}{4-x}` `(x ne -3; x ne 4)`

`⇔\frac{x+4}{x+3}+\frac{12x+7}{(x+3)(x-4)}+\frac{1-2x}{x-4}=0`

`⇔\frac{(x+4)(x-4)+12x+7+(1-2x)(x+3)}{(x+3)(x-4)}=0`

`⇒x^2-16+12x+7-2x^2-5x+3=0`

`⇔-x^2+7x-6=0`

`⇔x^2-7x+6=0`

`⇔x^2-x-6x+6=0`

`⇔x(x-1)-6(x-1)=0`

`⇔(x-6)(x-1)=0`

`⇔` \(\left[ \begin{array}{l}x-6=0\\x-1=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=6(tm)\\x=1(tm)\end{array} \right.\)