Đáp án:

$\begin{array}{l}
1)\dfrac{{x - 1}}{2} = \dfrac{{2x - 1}}{{ - 5}}\\
 \Rightarrow \left( { - 5} \right).\left( {x - 1} \right) = 2.\left( {2x - 1} \right)\\
 \Rightarrow  - 5x + 5 = 4x - 2\\
 \Rightarrow 4x + 5x = 5 + 2\\
 \Rightarrow 9x = 7\\
 \Rightarrow x = \dfrac{7}{9}\\
Vậy\,x = \dfrac{7}{9}\\
2)\dfrac{2}{3} + \dfrac{{x + 1}}{{ - 6}} = \dfrac{{2x - 1}}{2}\\
 \Rightarrow \dfrac{{2.2 - x - 1}}{6} = \dfrac{{3\left( {2x - 1} \right)}}{6}\\
 \Rightarrow 4 - x - 1 = 6x - 3\\
 \Rightarrow 6x + x = 4 - 1 + 3\\
 \Rightarrow 7x = 6\\
 \Rightarrow x = \dfrac{6}{7}\\
Vậy\,x = \dfrac{6}{7}\\
3){x^2} - 2x - 99 = 0\\
 \Rightarrow {x^2} - 11x + 9x - 99 = 0\\
 \Rightarrow \left( {x - 11} \right)\left( {x + 9} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 11\\
x =  - 9
\end{array} \right.\\
Vậy\,x =  - 9;x = 11\\
4)4{x^2} + 4x - 624 = 0\\
 \Rightarrow 4{x^2} + 4x + 1 - 625 = 0\\
 \Rightarrow \left( {2x + 1} \right) = {25^2}\\
 \Rightarrow \left[ \begin{array}{l}
2x + 1 = 25\\
2x + 1 =  - 25
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
2x = 24\\
2x =  - 26
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 12\\
x =  - 13
\end{array} \right.\\
Vậy\,x = 12;x =  - 13\\
5){x^3} - 3{x^2} + 3x - 9 = 0\\
 \Rightarrow {x^2}\left( {x - 3} \right) + 3\left( {x - 3} \right) = 0\\
 \Rightarrow \left( {{x^2} + 3} \right)\left( {x - 3} \right) = 0\\
 \Rightarrow x = 3\\
Vậy\,x = 3\\
6){x^3} + 3{x^2} + 3x + 28 = 0\\
 \Rightarrow {x^3} + 3{x^2} + 3x + 1 + 27 = 0\\
 \Rightarrow {\left( {x + 1} \right)^3} =  - 27\\
 \Rightarrow x + 1 =  - 3\\
 \Rightarrow x =  - 4\\
Vậy\,x =  - 4\\
7)\left| {2x - 5} \right| = 1\\
 \Rightarrow 2x - 5 = 1\,hoặc\,2x - 5 =  - 1\\
 \Rightarrow x = 3\,hoặc\,x = 2\\
Vậy\,x = 3\,hoặc\,x = 2\\
8)\left| {2x - 3} \right| + \left| {6 - 4x} \right| = 6\\
 \Rightarrow \left| {2x - 3} \right| + 2.\left| {2x - 3} \right| = 6\\
 \Rightarrow 3\left| {2x - 3} \right| = 6\\
 \Rightarrow \left| {2x - 3} \right| = 2\\
 \Rightarrow 2x - 3 = 2\,hoặc\,2x - 3 =  - 2\\
 \Rightarrow 2x = 5\,hoặc\,2x = 1\\
 \Rightarrow x = \dfrac{5}{2}\,hoặc\,x = \dfrac{1}{2}\\
Vậy\,x = \dfrac{5}{2}\,hoặc\,x = \dfrac{1}{2}\\
9)\left| {2x - 3} \right| = \left| {7 - x} \right|\\
 \Rightarrow \left[ \begin{array}{l}
2x - 3 = 7 - x\\
2x - 3 = x - 7
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
3x = 10\\
x =  - 4
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x \Rightarrow \dfrac{{10}}{3}\\
x =  - 4
\end{array} \right.
\end{array}$