Question

[Trâm Vũ]

Giải phương trình sau:

\(\frac{x+1}{2018}\)+\(\frac{x+2}{2019}\)=\(\frac{x+3}{2020}\)+\(\frac{x+4}{2021}\)

Answer 1

[nguyễn ngọc dinh]

\(\frac{x+1}{2018}+\frac{x+2}{2019}=\frac{x+3}{2020}+\frac{x+4}{2021}\)

\(\Leftrightarrow\left(\frac{x+1}{2018}-1\right)+\left(\frac{x+2}{2019}-1\right)=\left(\frac{x+3}{2020}-1\right)+\left(\frac{x+4}{2021}-1\right)\)

\(\Leftrightarrow\frac{x-2017}{2018}+\frac{x-2017}{2019}=\frac{x-2017}{2020}+\frac{x-2017}{2021}\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)

\(\Leftrightarrow x-2017=0\)\(\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)

\(\Leftrightarrow x=2017\)

Vậy \(S=\left\{2017\right\}\)