So sánh A và B biết A= 2024 / 2025+ 2-25 / 2026 + 2026/ 2024

\(A = \frac{{2024}}{{2025}} + \frac{{2025}}{{2026}} + \frac{{2026}}{{2024}}\)

\( = \left( {1 - \frac{1}{{2025}}} \right) + \left( {1 - \frac{1}{{2026}}} \right) + \left( {1 + \frac{2}{{2024}}} \right)\)

\( = 3 + \left( {\frac{2}{{2024}} - \frac{1}{{2025}} - \frac{1}{{2026}}} \right)\)

Ta thấy:

      \(\frac{2}{{2024}} > \frac{1}{{2025}} + \frac{1}{{2026}} \Rightarrow \left( {\frac{2}{{2024}} - \frac{1}{{2025}} - \frac{1}{{2026}}} \right) > 0\)

Suy ra \(A > 3\)

Với \(B = \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{{15}}\)

Nhóm như sau:

       \(\frac{1}{3} + \frac{1}{4} + \frac{1}{5} < \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\)

  \(\frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{{10}} + \frac{1}{{11}} < 6 \times \frac{1}{6} = 1\)

      \(\frac{1}{{12}} + \frac{1}{{13}} + \frac{1}{{14}} + \frac{1}{{15}} < 4 \times \frac{1}{{12}} < \frac{1}{3}\)

Vậy:

                  \(B < \frac{1}{2} + 1 + 1 + \frac{1}{3} = 2.833 \ldots < 3\)

Kết luận:

                                       \(A > 3 > B \Rightarrow A > B\)