Đáp án:

$\begin{array}{l}
1)7x - 4 = 3x + 12\\
 \Leftrightarrow 7x - 3x = 12 + 4\\
 \Leftrightarrow 4x = 16\\
 \Leftrightarrow x = 4\\
Vậy\,x = 4\\
 - 6x - 12 = 4x - 23\\
 \Leftrightarrow  - 6x - 4x =  - 23 + 12\\
 \Leftrightarrow  - 10x =  - 11\\
 \Leftrightarrow x = \dfrac{{11}}{{10}}\\
Vậy\,x = \dfrac{{11}}{{10}}\\
2)Dk:x \ne  - \dfrac{1}{2};x \ne \dfrac{1}{2}\\
\dfrac{2}{{2x + 1}} + \dfrac{x}{{4{x^2} - 1}} = \dfrac{7}{{2x - 1}}\\
 \Leftrightarrow \dfrac{{2.\left( {2x - 1} \right) + x}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}} = \dfrac{{7\left( {2x + 1} \right)}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}}\\
 \Leftrightarrow 4x - 2 + x = 14x + 7\\
 \Leftrightarrow 9x =  - 9\\
 \Leftrightarrow x =  - 1\left( {tm} \right)\\
Vậy\,x =  - 1\\
3)Dk:2 - x \ge 0 \Leftrightarrow x \le 2\\
\left| {2x - 3} \right| = 2 - x\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 2 - x\\
2x - 3 = x - 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = 5\\
x = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{3}\left( {tm} \right)\\
x = 1\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{5}{3};x = 1\\
\left| {4x - 3} \right| = 3 - 4x\\
 \Leftrightarrow 4x - 3 \le 0\\
 \Leftrightarrow 4x \le 3\\
 \Leftrightarrow x \le \dfrac{3}{4}\\
Vậy\,x \le \dfrac{3}{4}\\
\left| {3x} \right| - x = 4x + 5\\
 \Leftrightarrow \left| {3x} \right| = 5x + 5\left( {dk:x \ge  - 1} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = 5x + 5\\
3x =  - 5x - 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x =  - 5\\
8x =  - 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{5}{2}\left( {ktm} \right)\\
x =  - \dfrac{5}{8}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x =  - \dfrac{5}{8}
\end{array}$

Đáp án:

`1)` `7x - 4 = 3x + 12`

`<=>` `7x - 3x = 12 + 4`

`<=>` `4x = 16`

`<=>` `x = 4`

Vậy `:` `S = { 4 }`.

`-6x - 12 = 4x - 23`

`<=>` `-6x - 4x = -23 + 12`

`<=>` `-10x = -11`

`<=>` `x = 11/10`

Vậy `:` `S = { 11/10 }`.

`2)` `2 / ( 2x + 1 ) + x / ( 4x^2 - 1 ) = 7 / ( 2x - 1 )` `( x \ne -1/2 ; x \ne 1/2 )`

`<=>` `2 / ( 2x + 1 ) + x / ( 4x^2 - 1 ) - 7 / ( 2x - 1 ) = 0`

`<=>` `[ 2 ( 2x - 1 ) + x - 7 ( 2x + 1 ) ] / [ ( 2x + 1 ) ( 2x - 1 ) ] = 0`

`<=>` `4x - 2 + x - 14x - 7  = 0`

`<=>` `-9x - 9 = 0`

`<=>` `-9 ( x + 1 ) = 0`

`<=>` `x + 1 = 0`

`<=>` `x = -1`

Vậy `:` `S = { -1 }`.

`3)` `| 2x - 3 | = 2 - x` `( x <= 2 )`

`<=>` \(\left[ \begin{array}{l}2x - 3 = 2 - x\\2x - 3 = x - 2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}3x = 5\\x = 1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x = \dfrac{5}{3} ( N )\\x = 1 ( N )\end{array} \right.\) 

Vậy `:` `S = { 5/3 ; 1 }`.

`| 4x - 3 | = 3 - 4x`

`<=>` `4x - 3 <= 0`

`<=>` `4x <= 3`

`<=>` `x <= 3/4`

Vậy  `x <= 3/4`.

`| 3x | - x = 4x + 5`

`<=>` `| 3x | = 5x + 5` `( x >= -1 )`

`<=>` \(\left[ \begin{array}{l}3x = 5x + 5\\3x = -5x - 5\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x = -5\\8x = -5\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x = -\dfrac{5}{2} ( L )\\x = -\dfrac{5}{8} ( N )\end{array} \right.\) 

Vậy `:` `S = { -5/8 }`.