Giải thích các bước giải:

Theo đề ta có:

$|2x-1|\cdot (\dfrac1{2010}+\dfrac1{2011}+...+\dfrac1{2016})=\dfrac9{2010}+\dfrac8{2011}+...+\dfrac3{2016}+7$

$\to |2x-1|\cdot (\dfrac1{2010}+\dfrac1{2011}+...+\dfrac1{2016})=(1+\dfrac9{2010})+(1+\dfrac8{2011})+...+(1+\dfrac3{2016})$

$\to |2x-1|\cdot (\dfrac1{2010}+\dfrac1{2011}+...+\dfrac1{2016})=\dfrac{2019}{2010}+\dfrac{2019}{2011}+...+\dfrac{2019}{2016}$

$\to |2x-1|\cdot (\dfrac1{2010}+\dfrac1{2011}+...+\dfrac1{2016})=2019\cdot (\dfrac1{2010}+\dfrac1{2011}+...+\dfrac1{2016})$

$\to |2x-1|=2019$

$\to 2x-1=2019$ hoặc $2x-1=-2019$

$\to x=1010$ hoặc $x=-1009$