Tính a2023 – 2b2024.

A

\(f'\left( x \right) = \frac{{\frac{{2x}}{{{x^2} + 1}}.x - \ln \left( {{x^2} + 1} \right)}}{{{x^2}}}\)\( = \frac{{2{x^2} - \left( {{x^2} + 1} \right)\ln \left( {{x^2} + 1} \right)}}{{{x^2}\left( {{x^2} + 1} \right)}}\).

Khi đó \(f'\left( 1 \right) = \frac{{{{2.1}^2} - \left( {{1^2} + 1} \right)\ln \left( {{1^2} + 1} \right)}}{{{1^2}\left( {{1^2} + 1} \right)}} = 1 - \ln 2\).

Suy ra a = −1; b = 1. Do đó a²⁰²³ – 2b²⁰²⁴ = (−1)²⁰²³ – 2.1²⁰²⁴ = −3.