Ta có:

\(a=\frac{2021^{2022}-2018}{2021^{2022}-2020}=\frac{2021^{2022}-2020+2}{2021^{2022}-2020}=\frac{2021^{2022}-2020}{2021^{2022}-2020}+\frac{2}{2021^{2022}-2020}=1+\frac{2}{2021^{2022}-2020}\)

\(b=\frac{2021^{2022}-2020}{2021^{2022}-2022}=\frac{2021^{2022}-2022+2}{2021^{2022}-2022}=\frac{2021^{2022}-2022}{2021^{2022}-2022}+\frac{2}{2021^{2022}-2022}=1+\frac{2}{2021^{2022}-2022}\)

Ta thấy: \(2020<2021\rarr2021^{2022}-2020>2021^{2022}-2022\to\frac{2}{2021^{2022}-2020}<\frac{2}{2021^{2022}-2022}\to1+\frac{2}{2021^{2022}-2020}<1+\frac{2}{2021^{2022}-2022}\)

\(\to a<b\)

Vậy \(a<b\)

Mình gửi câu trả lời bằng hình ảnh dưới nhé