Question

Giải các phương trình:

\(\dfrac{x+1}{2019}+\dfrac{x+2}{2018}=\dfrac{x+2017}{3}+\dfrac{x+2016}{4}\)

Answer 1

\(\dfrac{x+1}{2019}+\dfrac{x+2}{2018}=\dfrac{x+2017}{3}+\dfrac{x+2016}{4}\)

\(\Leftrightarrow\left(\dfrac{x+1}{2019}+1\right)+\left(\dfrac{x+2}{2018}+1\right)=\left(\dfrac{x+2017}{3}+1\right)+\left(\dfrac{x+2016}{4}+1\right)\)

\(\Leftrightarrow\dfrac{x+2020}{2019}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{3}-\dfrac{x+2020}{4}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{3}-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x+2020=0\) ( do \(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{3}-\dfrac{1}{4}\ne0\))

\(\Leftrightarrow x=-2020\)

Vậy phương trình có tập nghiệm S = \(\left\{-2020\right\}\)