Ta có: \(\frac{x-5}{2016}+\frac{x-4}{2017}=\frac{x-3}{2018}+\frac{x-2}{2019}\)
\(\Leftrightarrow\frac{x-5}{2016}-1+\frac{x-4}{2017}-1=\frac{x-3}{2018}-1+\frac{x-2}{2019}-1\)
\(\Leftrightarrow\frac{x-2021}{2016}+\frac{x-2021}{2017}=\frac{x-2021}{2018}+\frac{x-2021}{2019}\)
\(\Leftrightarrow\frac{x-2021}{2016}+\frac{x-2021}{2017}-\frac{x-2021}{2018}-\frac{x-2021}{2019}=0\)
\(\Leftrightarrow\left(x-2021\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)
mà \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\ne0\)
nên x-2021=0
hay x=2021
Vậy: x=2021
Ta có:
\(\frac{x-5}{2016}+\frac{x-4}{2017}=\frac{x-3}{2018}+\frac{x-2}{2019}\)
\(\Leftrightarrow\left(\frac{x-5}{2016}-1\right)+\left(\frac{x-4}{2017}-1\right)=\left(\frac{x-3}{2018}-1\right)+\left(\frac{x-2}{2019}-1\right)\)
\(\Leftrightarrow\frac{x-2021}{2016}+\frac{x-2021}{2017}=\frac{x-2021}{2018}+\frac{x-2021}{2019}\)
\(\Leftrightarrow\frac{x-2021}{2016}+\frac{x-2021}{2017}-\frac{x-2021}{2018}-\frac{x-2021}{2019}=0\)
\(\Leftrightarrow\left(x-2021\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)
Vì \(\frac{1}{2016}>\frac{1}{2018}\) và \(\frac{1}{2017}>\frac{1}{2019}\) nên \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)
\(\Rightarrow x-2021=0\)
\(\Rightarrow x=2021\)
Vậy phương trình có nghiệm duy nhất là \(x=2021\)
