Giải thích các bước giải:

Bài 1:

Ta có:

$2025\cdot 2022=(2024+1)(2023-1)=2024\cdot 2023-2024+2023-1 =2024\cdot 2023-2<2024\cdot 2023$

$\to \sqrt{2025}\cdot \sqrt{2022}<\sqrt{2024}\cdot \sqrt{2023}$

$\to 2\sqrt{2025}\cdot \sqrt{2022}<2\sqrt{2024}\cdot \sqrt{2023}$

$\to 4047+2\sqrt{2025}\cdot \sqrt{2022}<4047+2\sqrt{2024}\cdot \sqrt{2023}$

$\to 2025+2\sqrt{2025}\cdot \sqrt{2022}+2022<2024+2\sqrt{2024}\cdot \sqrt{2023}+2023$

$\to (\sqrt{2025}+\sqrt{2022})^2<(\sqrt{2024}+\sqrt{2023})^2$

$\to \sqrt{2025}+\sqrt{2022}<\sqrt{2024}+\sqrt{2023}$

$\to \sqrt{2025}-\sqrt{2023}<\sqrt{2024}-\sqrt{2022}$

$\to A<B$

Bài 2:

Ta có: 

$\dfrac1{\sqrt{2k-1}+\sqrt{2k+1}}=\dfrac12\cdot \dfrac2{\sqrt{2k-1}+\sqrt{2k+1}}=\dfrac12\cdot \dfrac{(2k+1)-(2k-1)}{\sqrt{2k-1}+\sqrt{2k+1}}=\dfrac12\cdot \dfrac{(\sqrt{2k-1}+\sqrt{2k+1})(\sqrt{2k+1}-\sqrt{2k-1})}{\sqrt{2k-1}+\sqrt{2k+1}}=\dfrac12(-\sqrt{2k-1}+\sqrt{2k+1})$

Áp dụng vào tính S:

$S=\dfrac12(-\sqrt1+\sqrt3)+\dfrac12(-\sqrt3+\sqrt5)+....+\dfrac12(-\sqrt{77}+\sqrt{79})+\dfrac12(-\sqrt{79}+\sqrt{81})$

$\to S=\dfrac12(-\sqrt1+\sqrt3-\sqrt3+\sqrt5-....-\sqrt{77}+\sqrt{79}-\sqrt{79}+\sqrt{81})$

$\to S=\dfrac12(-\sqrt1+\sqrt{81})$

$\to S=4$