Giải thích các bước giải:
Ta có :
$(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015})x=\dfrac{2014}{1}+\dfrac{2013}{2}+..+\dfrac{2}{2013}+\dfrac{1}{2014}$
$\to (\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015})x=2014+\dfrac{2013}{2}+..+\dfrac{2}{2013}+\dfrac{1}{2014}$
$\to (\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015})x=1+(\dfrac{2013}{2}+1)+..+(\dfrac{2}{2013}+1)+(\dfrac{1}{2014}+1)$
$\to (\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015})x=1+(\dfrac{2013}{2}+1)+..+(\dfrac{2}{2013}+1)+(\dfrac{1}{2014}+1)$
$\to (\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015})x=1+\dfrac{2015}{2}+\dfrac{2015}{3}+..+\dfrac{2015}{2014}$
$\to (\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015})x=\dfrac{2015}{2}+\dfrac{2015}{3}+..+\dfrac{2015}{2014}+\dfrac{2015}{2015}$
$\to (\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015})x=2015(\dfrac{1}{2}+..+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015})$
$\to x=2015$
