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Lời giải:
$a=1+5+5^2+5^3+...+5^{2022}+5^{2023}$
$5a=5+5^2+5^3+5^4+....+5^{2023}+5^{2024}$
$\Rightarrow 5a-a=5^{2024}-1$
$\Rightarrow 4a=5^{2024}-1$
$\Rightarrow 4a+1=5^{2024}\vdots 5^{2023}$ (đpcm)
A = \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2022}{50^8}\)
A = \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2021}{50^8}\) + \(\dfrac{1}{50^8}\)
B = \(\dfrac{2023}{50^{10}}\) + \(\dfrac{2021}{5^8}\) = \(\dfrac{2022}{50^{10}}\) + \(\dfrac{1}{50^{10}}\) + \(\dfrac{2021}{50^8}\)
Vì: \(\dfrac{1}{50^{10}}\) < \(\dfrac{1}{50^8}\) nên \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2021}{50^8}\) + \(\dfrac{1}{50^{10}}\) < \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2021}{50^8}\) + \(\dfrac{1}{50^8}\)
Vậy A > B
A = \(\dfrac{2023}{2022^2+1}\) + \(\dfrac{2023}{2022^2+2}\) + ... + \(\dfrac{2023}{2022^2+3}\)+.... + \(\dfrac{2023}{2022^2+2022}\)
A = 2023.(\(\dfrac{1}{2022^2+1}\) + \(\dfrac{1}{2022^2+2}\) + ... + \(\dfrac{1}{2022^2+2022}\))
\(\dfrac{1}{2022^2+1}\) > \(\dfrac{1}{2022^2+2}\) > .... > \(\dfrac{1}{2022^2+2022}\)
Vì dãy phân số trên có 2022 phân số nên:
A > 2023. \(\dfrac{1}{2022^2+2022}\). 2022
A > 2023. \(\dfrac{2022}{2022^2+2022}\)
A > 2023. \(\dfrac{2022}{2022.\left(2022+1\right)}\)
A > \(\dfrac{2023.2022}{2022.2023}\) = 1
A > 1 (đpcm)