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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: \(A>B\)
Áp dụng tính chất : Nếu \(\dfrac{a}{b}\) < 1 thì \(\dfrac{a}{b}\) < \(\dfrac{a+n}{b+n}\) ( a ϵ N; b; n ϵ N* )
Ta có \(B=\dfrac{10^{2021}+1}{10^{2022}+1}< \dfrac{10^{2021}+10}{10^{2022}+10}=\dfrac{10\left(10^{2020}+1\right)}{10\left(10^{2021}+1\right)}=\dfrac{10^{2020}+1}{10^{2021}+1}=A\)
Vậy A > B
A = \(\dfrac{10^{2020}+1}{10^{2021}+1}\) ⇒ 10\(\times\) A = \(\dfrac{10^{2020}+1}{10^{2021}+1}\) \(\times\) 10
10A = \(\dfrac{10^{2021}+10}{10^{2021}+1}\) =1+\(\dfrac{9}{10^{2021}+1}\)
B = \(\dfrac{10^{2021}+1}{10^{2022}+1}\) ⇒ 10 \(\times\) B = \(\dfrac{10^{2021}+1}{10^{2022}+1}\) \(\times\) 10
10B = \(\dfrac{10^{2022}+10}{10^{2022}+1}\) = 1 + \(\dfrac{9}{10^{2022}+1}\)
Vì \(\dfrac{9}{10^{2021}+1}\) > \(\dfrac{9}{10^{2022}+1}\)
Vậy 10A > 10B ⇒ A > B
\(C=\dfrac{10^{2021}+10-9}{10^{2020}+1}=10-\dfrac{9}{10^{2020}+1}\)
\(D=\dfrac{10^{2022}+10-9}{10^{2021}+1}=10-\dfrac{9}{10^{2021}+1}\)
mà \(10^{2020}+1< 10^{2021}+1\)
nên \(-\dfrac{9}{10^{2020}+1}< -\dfrac{9}{10^{2021}+1}\)
hay C<D
Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)
=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)
Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)
=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)
Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)
=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)
=> 10B < 10A
=> B < A
b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)
Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)
=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> B < A
\(\dfrac{1}{10}A=\dfrac{10^{2023}+5}{10^{2023}+50}=1-\dfrac{45}{10^{2023}+50}\)
\(\dfrac{1}{10}B=\dfrac{10^{2022}+5}{10^{2022}+50}=1-\dfrac{45}{10^{2022}+50}\)
10^2023+50>10^2022+50
=>-45/10^2023+50<-45/10^2020+50
=>1/10A<1/10B
=>A<B
\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)
\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
2023>2022
=>10^2023+1>10^2022+1
=>10A<10B
=>A<B
\(2.A=\frac{2^{2021}-2}{2^{2021}-1}=1-\frac{1}{2^{2021}-1}\)
\(2B=\frac{2^{2022}-2}{2^{2022}-1}=1-\frac{1}{2^{2022}-1}\)
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
\(A=\dfrac{2021^{10}-2021+2020}{2021^9-1}\\ =\dfrac{2021\left(2021^9-1\right)+2020}{2021^9-1}\\ =2021+\dfrac{2020}{2021^9-1}\\ B=\dfrac{2021^{11}-1}{2021^{10}-1}=2021+\dfrac{2020}{2021^{10}-1}\)
Ta có:
\(2021^9-1< 2021^{10}-1\\ \Rightarrow\dfrac{2020}{2021^9-1}>\dfrac{2020}{2021^{10}-1}\)
Do đó 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 $A> B$