Đáp án:

$\begin{array}{l}
a)\dfrac{1}{4}x - \dfrac{1}{3} =  - \dfrac{5}{9}\\
 \Rightarrow \dfrac{1}{4}x =  - \dfrac{5}{9} + \dfrac{1}{3}\\
 \Rightarrow \dfrac{1}{4}x =  - \dfrac{5}{9} + \dfrac{3}{9} =  - \dfrac{2}{9}\\
 \Rightarrow x = \dfrac{{ - 2}}{9}:\dfrac{1}{4}\\
 \Rightarrow x = \dfrac{{ - 8}}{9}\\
b)\dfrac{{11}}{{12}} - \left( {\dfrac{2}{5} - x} \right) = \dfrac{3}{4}\\
 \Rightarrow \dfrac{{11}}{{12}} - \dfrac{2}{5} + x = \dfrac{3}{4}\\
 \Rightarrow x = \dfrac{3}{4} - \dfrac{{11}}{{12}} + \dfrac{2}{5}\\
 \Rightarrow x = \dfrac{{3.15 - 5.11 + 2.12}}{{60}}\\
 \Rightarrow x = \dfrac{{14}}{{60}} = \dfrac{7}{{30}}\\
c)2007,5 - \left| {x - 1,5} \right| = 0\\
 \Rightarrow \left| {x - 1,5} \right| = 2007,5\\
 \Rightarrow \left[ \begin{array}{l}
x - 1,5 = 2007,5\\
x - 1,5 =  - 2007,5
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 2009\\
x =  - 2006
\end{array} \right.\\
d){\left( {x - 2} \right)^{2012}} + {\left| {{y^2} - 9} \right|^{2014}} = 0\\
 \Rightarrow \left\{ \begin{array}{l}
x - 2 = 0\\
{y^2} - 9 = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 2\\
{y^2} = 9
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = 3\\
y =  - 3
\end{array} \right.\\
e)\dfrac{{45 - x}}{{1963}} + \dfrac{{40 - x}}{{1968}} + \dfrac{{35 - x}}{{1973}} + \dfrac{{30 - x}}{{1978}} + 4 = 0\\
 \Rightarrow \dfrac{{45 - x}}{{1963}} + 1 + \dfrac{{40 - x}}{{1968}} + 1\\
 + \dfrac{{35 - x}}{{1973}} + 1 + \dfrac{{30 - x}}{{1978}} + 1 = 0\\
 \Rightarrow \dfrac{{45 - x + 1963}}{{1963}} + \dfrac{{40 - x + 1968}}{{1968}}\\
 + \dfrac{{35 - x + 1973}}{{1973}} + \dfrac{{30 - x + 1978}}{{1978}} = 0\\
 \Rightarrow \dfrac{{2008 - x}}{{1963}} + \dfrac{{2008 - x}}{{1968}} + \dfrac{{2008 - x}}{{1973}} + \dfrac{{2008 - x}}{{1978}} = 0\\
 \Rightarrow \left( {2008 - x} \right).\left( {\dfrac{1}{{1963}} + \dfrac{1}{{1968}} + \dfrac{1}{{1973}} + \dfrac{1}{{1978}}} \right) = 0\\
 \Rightarrow x = 2008\\
f)x - \dfrac{{20}}{{11.13}} - \dfrac{{20}}{{13.15}} - \dfrac{{20}}{{15.17}} - ...\dfrac{{20}}{{53.55}} = \dfrac{3}{{11}}\\
 \Rightarrow x - 10.\left( {\dfrac{2}{{11.13}} + \dfrac{2}{{13.15}} + ... + \dfrac{1}{{53.55}}} \right) = \dfrac{3}{{11}}\\
 \Rightarrow x - 10.\left( {\dfrac{1}{{11}} - \dfrac{1}{{55}}} \right) = \dfrac{3}{{11}}\\
 \Rightarrow x - 10.\dfrac{4}{{55}} = \dfrac{3}{{11}}\\
 \Rightarrow x = \dfrac{3}{{11}} + \dfrac{{40}}{{55}} = \dfrac{{55}}{{55}} = 1
\end{array}$