Ta có: \(\dfrac{2x-5}{2015}+\dfrac{2x-6}{2016}=\dfrac{2x+3}{2007}+\dfrac{2x+4}{2006}\)
\(\Leftrightarrow\dfrac{2x-5}{2015}+1+\dfrac{2x-6}{2016}+1=\dfrac{2x+3}{2007}+1+\dfrac{2x+4}{2006}+1\)
\(\Leftrightarrow\dfrac{2x+2010}{2015}+\dfrac{2x+2010}{2016}-\dfrac{2x+2010}{2007}-\dfrac{2x+2010}{2006}=0\)
\(\Leftrightarrow\left(2x+2010\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
mà \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)
nên 2x+2010=0
\(\Leftrightarrow2x=-2010\)
hay x=-1005
Vậy: S={-1005}
\(\dfrac{2x-5}{2015}+\dfrac{2x-6}{2016}=\dfrac{2x+3}{2007}+\dfrac{2x+4}{2006}\)
= \(\left(\dfrac{2x-5}{2015}-1\right)+\left(\dfrac{2x-6}{2016}-1\right)=\left(\dfrac{2x+3}{2007}+1\right)+\left(\dfrac{2x+4}{2006}+1\right)\)
=\(\dfrac{2x-2010}{2015}+\dfrac{2x-2010}{2016}=\dfrac{2x+2010}{2007}+\dfrac{2x+2010}{2006}\)
=\(\dfrac{2x-2010}{2015}+\dfrac{2x-2010}{2016}-\dfrac{2x-2010}{2007}-\dfrac{2x-2010}{2006}\)=0
=(2x-2010)\(\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)\)=0
vì \(\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)\)≠0
⇒ 2x-2010=0
⇔2x=2010
⇔x=1005
