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Ta có:
\(10A=\dfrac{10\left(10^{2020}+1\right)}{10^{2021}+1}=\dfrac{10^{2021}+10}{10^{2021}+1}=1+\dfrac{9}{10^{2021}+1}\)
\(10B=\dfrac{10\left(10^{2021}+1\right)}{10^{2022}+1}=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
⇒ \(10A>10B\) ( vì \(\dfrac{9}{10^{2021}+1}>\dfrac{9}{10^{2022}+1}\) )
Suy ra: \(A>B\)
2020/2021<1
2021/2022<1
2022/2023<1
2023/2020=1+1/2020+1/2020+1/2020>1+1/2021+1/2022+1/2023
=>B>2020/2021+2021/2022+2022/2023+1/2021+1/2022+1/2023+1=4
\(2.A=\frac{2^{2021}-2}{2^{2021}-1}=1-\frac{1}{2^{2021}-1}\)
\(2B=\frac{2^{2022}-2}{2^{2022}-1}=1-\frac{1}{2^{2022}-1}\)
dó \(\frac{1}{2^{2022}-1}< \frac{1}{2^{2021}-1}\Rightarrow1-\frac{1}{2^{2022}-1}>1-\frac{1}{2^{2021}-1}\Rightarrow A< B\)
HT
\(\dfrac{-11}{-32}>\dfrac{16}{49}\)
\(\dfrac{-2020}{-2021}>\dfrac{-2021}{2022}\)
Lời giải:
$6A=\frac{6^{2021}+6}{6^{2021}+1}=1+\frac{5}{6^{2021}+1}>1+\frac{5}{6^{2022}+1}$
$=\frac{6^{2022}+6}{6^{2022}+1}=6.\frac{6^{2021}+1}{6^{2022}+1}=6B$
$\Rightarrow A>B$
Lời giải:
$10A=\frac{10^{2021}-10}{10^{2021}-1}=\frac{10^{2021}-1-9}{10^{2021}-1}$
$=1-\frac{9}{10^{2021}-1}>1$
$10B=\frac{10^{2022}+10}{10^{2022}+1}=\frac{10^{2022}+1+9}{10^{2022}+1}$
$=1+\frac{9}{10^{2022}+1}<1$
$\Rightarrow 10A> 1> 10B$
Suy ra $A> B$
2023/2022=1+1/2022
2022/2021=1+1/2021
mà 2022>2021
nên 2023/2022<2022/2021
a: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)
mà 17^19+1>17^18+1
nên A<B
b: \(2C=\dfrac{2^{2021}-2}{2^{2021}-1}=1-\dfrac{1}{2^{2021}-1}\)
\(2D=\dfrac{2^{2022}-2}{2^{2022}-1}=1-\dfrac{1}{2^{2022}-1}\)
2^2021-1<2^2022-1
=>1/2^2021-1>1/2^2022-1
=>-1/2^2021-1<-1/2^2022-1
=>C<D
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
2022/2021 = 1 + 1/2021
2021/2020 = 1 + 1/2020
Do 2021 > 2020 nên 1/2021 < 1/2020
⇒ 1 + 1/2021 < 1 + 1/2020
⇒ 2022/2021 < 2021/2020
Có:
\(\dfrac{2022}{2021}-1=\dfrac{2022-2021}{2021}=\dfrac{1}{2021}\)
\(\dfrac{2021}{2020}-1=\dfrac{2021-2020}{2020}=\dfrac{1}{2020}\)
Ta thấy: \(2021>2020\)\(\Rightarrow\dfrac{1}{2021}< \dfrac{1}{2020}\)
hay \(\dfrac{2022}{2021}-1< \dfrac{2021}{2020}-1\Rightarrow\dfrac{2022}{2021}< \dfrac{2021}{2020}\)