\[\begin{array}{l}
f)\left( {2x - 1} \right)\left( {x + \frac{2}{3}} \right) = 0\\
= > \left[ \begin{array}{l}
2x - 1 = 0\\
x + \frac{2}{3} = 0
\end{array} \right.\\
= > \left[ \begin{array}{l}
x = \frac{1}{2}\\
x = \frac{{ - 2}}{3}
\end{array} \right.\\
vay:S = \{ \frac{1}{2};\frac{{ - 2}}{3}\} \\
g)\frac{{x + 4}}{{2008}} + \frac{{x + 3}}{{2009}} = \frac{{x + 2}}{{2010}} + \frac{{x + 1}}{{2011}}\\
= > \frac{{x + 4}}{{2008}} + 1 + \frac{{x + 3}}{{2009}} + 1 = \frac{{x + 2}}{{2010}} + 1 + \frac{{x + 1}}{{2011}} + 1\\
= > \frac{{x + 2012}}{{2008}} + \frac{{x + 2012}}{{2009}} = \frac{{x + 2012}}{{2010}} + \frac{{x + 2012}}{{2011}}\\
= > \left( {x + 2012} \right)\left( {\frac{1}{{2008}} + \frac{1}{{2009}} - \frac{1}{{2010}} - \frac{1}{{2011}}} \right) = 0\\
= > x + 2012 = 0\left( {do:\frac{1}{{2008}} + \frac{1}{{2009}} - \frac{1}{{2010}} - \frac{1}{{2011}} \ne 0} \right)\\
= > x = - 2012\\
Vay:x = - 2012
\end{array}\]