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\(D=\frac{1006-\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2011}\right)}{\frac{2}{3}+\frac{4}{5}+...+\frac{2010}{2011}}\)
\(D=\frac{1006-\left(1+\frac{1}{3}+...+\frac{1}{2011}\right)}{1-\frac{1}{3}+1-\frac{1}{5}+...+1-\frac{1}{2011}}\)
\(D=\frac{1006-\left(1+\frac{1}{3}+...+\frac{1}{2011}\right)}{1006-\left(1+\frac{1}{3}+...+\frac{1}{2011}\right)}=1\)
A= \(1-\frac{2011}{2012}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\)
B=\(\left(\frac{2012}{1}-1\right)+\left(\frac{2012}{2}-1\right)+...+\left(\frac{2012}{2011}-1\right)\)
= \(\frac{2012}{1}-\frac{2012}{2012}+\frac{2012}{2}-\frac{2012}{2012}+...+\frac{2012}{2011}-\frac{2012}{2012}\)
=\(2012\left(1-\frac{1}{2012}+\frac{1}{2}-\frac{1}{2012}+...+\frac{1}{2011}-\frac{1}{2012}\right)\)
\(\Rightarrow\)\(\frac{B}{A}\)=\(\frac{2012\left(1-\frac{2011}{2012}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}{1-\frac{2011}{2012}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}\)= 2012
\(|x-\frac{1}{3}|=|\left(-3.2\right)+\frac{2}{5}|\)
\(\Rightarrow|x-\frac{1}{3}|=|-3.2+0.4|\)
\(\Rightarrow|x-\frac{1}{3}|=|-2.8|\)
\(\Rightarrow|x-\frac{1}{3}|=2.8\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2.8\\x-\frac{1}{3}=-2.8\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{43}{15}\\x=-\frac{41}{15}\end{cases}}\)
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