\[\begin{array}{l}
a)\,\,\,\,\,\frac{{x + 1}}{{10}} + \frac{{x + 1}}{{11}} = \frac{{x + 1}}{{12}} + \frac{{x + 1}}{{13}}\\
= > (x + 1)\left( {\frac{1}{{10}} + \frac{1}{{11}}} \right) = (x + 1)\left( {\frac{1}{{12}} + \frac{1}{{13}}} \right)\\
= > (x + 1)\left( {\frac{1}{{10}} + \frac{1}{{11}}} \right) - (x + 1)\left( {\frac{1}{{12}} + \frac{1}{{13}}} \right) = 0\\
= > (x + 1)\left( {\frac{1}{{10}} + \frac{1}{{11}} - \frac{1}{{12}} - \frac{1}{{13}}} \right) = 0\\
Vi\,\,\frac{1}{{10}} + \frac{1}{{11}} - \frac{1}{{12}} - \frac{1}{{13}} \ne 0 = > x + 1 = 0\,\,hay\,\,x = - 1\\
b)\,\,\,\,\frac{{x + 1}}{{2009}} + \frac{{x + 2}}{{2008}} + \frac{{x + 3}}{{2007}} = \frac{{x + 4}}{{2006}} + \frac{{x + 5}}{{2005}} + \frac{{x + 6}}{{2004}}\\
= > \,\frac{{x + 1}}{{2009}} + 1 + \frac{{x + 2}}{{2008}} + 1 + \frac{{x + 3}}{{2007}} + 1 = \frac{{x + 4}}{{2006}} + 1 + \frac{{x + 5}}{{2005}} + 1 + \frac{{x + 6}}{{2004}} + 1\\
= > \,\frac{{x + 2010}}{{2009}} + \frac{{x + 2010}}{{2008}} + \frac{{x + 2010}}{{2007}} = \frac{{x + 2010}}{{2006}} + \frac{{x + 2010}}{{2005}} + \frac{{x + 2010}}{{2004}}\\
Phan\,\,con\,\,lai\,\,tuong\,\,tu\,\,nhu\,\,bai\,\,a
\end{array}\]
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