b: \(10A=\dfrac{10^{1001}+10}{10^{1001}+1}=1+\dfrac{9}{10^{1001}+1}\)
\(10B=\dfrac{10^{1002}+10}{10^{1002}+1}=1+\dfrac{9}{10^{1002}+1}\)
Ta có: \(10^{1001}+1< 10^{1002}+1\)
=>\(\dfrac{9}{10^{1001}+1}>\dfrac{9}{10^{1002}+1}\)
=>\(1+\dfrac{9}{10^{1001}+1}>1+\dfrac{9}{10^{1002}+1}\)
=>10A>10B
=>A>B
c: \(\dfrac{1}{10}A=\dfrac{10^{2024}+1}{10^{2024}+10}=1-\dfrac{9}{10^{2024}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{2023}+1}{10^{2023}+10}=1-\dfrac{9}{10^{2023}+10}\)
Ta có: \(10^{2024}+10>10^{2023}+10\)
=>\(\dfrac{9}{10^{2024}+10}< \dfrac{9}{10^{2023}+10}\)
=>\(-\dfrac{9}{10^{2024}+10}>-\dfrac{9}{10^{2023}+10}\)
=>\(-\dfrac{9}{10^{2024}+10}+1>-\dfrac{9}{10^{2023}+10}+1\)
=>\(\dfrac{A}{10}>\dfrac{B}{10}\)
=>A>B
d: \(\dfrac{1}{10}A=\dfrac{10^{1000}+1}{10^{1000}+10}=1-\dfrac{9}{10^{1000}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{1001}+1}{10^{1001}+10}=1-\dfrac{9}{10^{1001}+10}\)
Ta có: \(10^{1000}+10< 10^{1001}+10\)
=>\(\dfrac{9}{10^{1000}+10}>\dfrac{9}{10^{1001}+10}\)
=>\(-\dfrac{9}{10^{1000}+10}< -\dfrac{9}{10^{1001}+10}\)
=>\(-\dfrac{9}{10^{1000}+10}+1< -\dfrac{9}{10^{1001}+10}+1\)
=>A/10<B/10
=>A<B
(đề bài: tìm x biết)
