So sánh A và B:
A= \(\dfrac{10^{2020}+1}{10^{2021}+1}\) B=\(\dfrac{10^{2021}+1}{10^{2022}+1}\)
Giúp mình với!
K
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AH
Akai Haruma
Giáo viên
30 tháng 4 2023
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 $A> B$
1 tháng 5 2023
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
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 5 2023
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
KJ
1
8 tháng 1 2022
\(C=\dfrac{10^{2021}+10-9}{10^{2020}+1}=10-\dfrac{9}{10^{2020}+1}\)
\(D=\dfrac{10^{2022}+10-9}{10^{2021}+1}=10-\dfrac{9}{10^{2021}+1}\)
mà \(10^{2020}+1< 10^{2021}+1\)
nên \(-\dfrac{9}{10^{2020}+1}< -\dfrac{9}{10^{2021}+1}\)
hay C<D
NA
Ngoc Anh Thai
Giáo viên
8 tháng 5 2021
\(A=\dfrac{2021^{10}-2021+2020}{2021^9-1}\\ =\dfrac{2021\left(2021^9-1\right)+2020}{2021^9-1}\\ =2021+\dfrac{2020}{2021^9-1}\\ B=\dfrac{2021^{11}-1}{2021^{10}-1}=2021+\dfrac{2020}{2021^{10}-1}\)
Ta có:
\(2021^9-1< 2021^{10}-1\\ \Rightarrow\dfrac{2020}{2021^9-1}>\dfrac{2020}{2021^{10}-1}\)
Do đó A > B.
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 4 2023
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
XO
30 tháng 7 2020
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
BD
16 tháng 7 2023
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
16 tháng 2 2023
\(\dfrac{1}{10}A=\dfrac{10^{2023}+5}{10^{2023}+50}=1-\dfrac{45}{10^{2023}+50}\)
\(\dfrac{1}{10}B=\dfrac{10^{2022}+5}{10^{2022}+50}=1-\dfrac{45}{10^{2022}+50}\)
10^2023+50>10^2022+50
=>-45/10^2023+50<-45/10^2020+50
=>1/10A<1/10B
=>A<B
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: \(A>B\)