Đáp án:
\(\begin{array}{l}
1e)\dfrac{{70}}{{29}}\\
2g)x = - 2012
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1e)2 + \dfrac{1}{{2 + \dfrac{1}{{2 + \dfrac{1}{{2 + \dfrac{1}{2}}}}}}} = 2 + \dfrac{1}{{2 + \dfrac{1}{{2 + \dfrac{1}{{\dfrac{5}{2}}}}}}}\\
= 2 + \dfrac{1}{{2 + \dfrac{1}{{2 + \dfrac{2}{5}}}}} = 2 + \dfrac{1}{{2 + \dfrac{1}{{\dfrac{{12}}{5}}}}}\\
= 2 + \dfrac{1}{{2 + \dfrac{5}{{12}}}} = 2 + \dfrac{1}{{\dfrac{{29}}{{12}}}} = 2 + \dfrac{{12}}{{29}}\\
= \dfrac{{70}}{{29}}\\
2g)\dfrac{{x + 4}}{{2008}} + 1 + \dfrac{{x + 3}}{{2009}} + 1 = \dfrac{{x + 2}}{{2010}} + 1 + \dfrac{{x + 1}}{{2011}} + 1\\
\to \dfrac{{x + 2012}}{{2008}} + \dfrac{{x + 2012}}{{2009}} = \dfrac{{x + 2012}}{{2010}} + \dfrac{{x + 2012}}{{2011}}\\
\to \left( {x + 2012} \right)\left( {\dfrac{1}{{2008}} + \dfrac{1}{{2009}} - \dfrac{1}{{2010}} - \dfrac{1}{{2011}}} \right) = 0\\
\to x + 2012 = 0\\
\to x = - 2012
\end{array}\)
