\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
\(\Leftrightarrow\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-1}{1008}-2+\dfrac{x}{2017}-1\) \(\Leftrightarrow\dfrac{x-3-2014}{2014}+\dfrac{x-2-2015}{2015}=\dfrac{x-1-2016}{1008}-\dfrac{x-2017}{2017}\) \(\Leftrightarrow\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{1008}+\dfrac{x-2017}{2017}\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\right)=0\)
Vì: \(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\ne0\)
Suy ra: x -2017 = 0
=> x = 2017
\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
⇔ \(\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-1}{2008}-2+\dfrac{x}{2017}-1\)
⇔\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2008}+\dfrac{x-2017}{2017}\)
⇔\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2008}-\dfrac{x-2017}{2017}=0\)
⇔\(\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2008}-\dfrac{1}{2017}\right)=0\)
⇔x-2017=0
⇔x=2017
vậy phương trình có tập nghiệm là S={2017}
\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-1}{1008}+\dfrac{x}{2017}-1\)
\(\Leftrightarrow\) \(\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-1}{1008}-2+\dfrac{x}{2017}-1\)
\(\Leftrightarrow\) \(\dfrac{x-3}{2014}-\dfrac{2014}{2014}+\dfrac{x-2}{2015}-\dfrac{2015}{2015}=\dfrac{x-1}{1008}-\dfrac{2016}{1008}+\dfrac{x}{2017}-\dfrac{2017}{2017}\)
\(\Leftrightarrow\)\(\dfrac{x-3-2014}{2014}+\dfrac{x-2-2015}{2015}=\dfrac{x-1-2016}{1008}+\dfrac{x-2017}{2017}\)
\(\Leftrightarrow\)\(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{1008}+\dfrac{x-2017}{2017}\)
\(\Leftrightarrow\) \(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{1008}-\dfrac{x-2017}{2017}=0\)
\(\Leftrightarrow\) \(\left(x-2017\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{1008}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow\) x - 2017 = 0
\(\Leftrightarrow\) x = 2017
Vậy.............
