Question

x+1 x+2 x+3 x+4 ------ + ---------- + --------- +--------=0 222 221 220 219

Answer 1

Sửa đề:\(\dfrac{x+1}{2022}+\dfrac{x+2}{2021}-\dfrac{x+3}{2020}-\dfrac{x+4}{2019}=0\)

\(\Leftrightarrow\left(\dfrac{x+1}{2022}+1\right)+\left(\dfrac{x+2}{2021}+1\right)-\left(\dfrac{x+3}{2020}+1\right)-\left(\dfrac{x+4}{2019}+1\right)=0\)

=>x+2023=0

=>x=-2023