Đáp án:

$2)\\ a)2021^{2021}+2021^{2020}<2022^{2021}\\ b)A<B.$

Giải thích các bước giải:

$2)\\ a)2021^{2021}+2021^{2020}\\ =2021^{2020}(2021+1)\\ =2022.2021^{2020}\\ 2022^{2021}=2022.2022^{2020}\\ \text{Do } 2022.2021^{2020}<2022.2022^{2020}\\ \Rightarrow 2021^{2021}+2021^{2020}<2022^{2021}\\ b)A=\dfrac{3^{2020}-1}{3^{2021}-1}\\ 3A=\dfrac{3^{2021}-3}{3^{2021}-1}=\dfrac{3^{2021}-1-2}{3^{2021}-1}=1-\dfrac{2}{3^{2021}-1}\\ B=\dfrac{3^{2021}-1}{3^{2022}-1}\\ 3B=\dfrac{3^{2022}-3}{3^{2022}-1}=\dfrac{3^{2022}-1-2}{3^{2022}-1}=1-\dfrac{2}{3^{2022}-1}\\ 3^{2021}-1<3^{2022}-1\\ \Rightarrow \dfrac{2}{3^{2021}-1}>\dfrac{2}{3^{2022}-1}\\ \Rightarrow 1-\dfrac{2}{3^{2021}-1}<1-\dfrac{2}{3^{2022}-1}\\ \Leftrightarrow 3A<3B\\ \Leftrightarrow A<B.$