a, ĐK: \(x+1\ge0\Leftrightarrow x\ge-1\)
Ta có: |3-2x|=x+1
=>\(\orbr{\begin{cases}3-2x=x+1\\3-2x=-x-1\end{cases}\Rightarrow\orbr{\begin{cases}x+2x=3-1\\-x+2x=3+1\end{cases}\Rightarrow}\orbr{\begin{cases}3x=2\\x=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\left(tmđk\right)\\x=4\left(tmđk\right)\end{cases}}}\)
Vậy x=2/3 hoặc x=4
b, Xét VP ta có: \(\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}=2013+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(=1+\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{2}{2012}\right)+\left(1+\frac{1}{2013}\right)\)
\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2012}+\frac{2014}{2013}+1\)
\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}=2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)\)
=>\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)\)
=>\(x=\frac{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}=2014\)
Vậy x=2014
