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16 tháng 5 2022

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:  \(A>B\)

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

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 

\(C=\dfrac{10^{2021}+10-9}{10^{2020}+1}=10-\dfrac{9}{10^{2020}+1}\)

\(D=\dfrac{10^{2022}+10-9}{10^{2021}+1}=10-\dfrac{9}{10^{2021}+1}\)

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.

 

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

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

13 tháng 2 2022

sai rồi

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.

 

\(\dfrac{1}{10}A=\dfrac{10^{2023}+5}{10^{2023}+50}=1-\dfrac{45}{10^{2023}+50}\)

\(\dfrac{1}{10}B=\dfrac{10^{2022}+5}{10^{2022}+50}=1-\dfrac{45}{10^{2022}+50}\)

10^2023+50>10^2022+50

=>-45/10^2023+50<-45/10^2020+50

=>1/10A<1/10B

=>A<B

16 tháng 2 2023

A phải lớn hơn B vì phần bù của số nào nhỏ hơn thì số đó lớn hơn bạn nhé. Nhưng dù sao cx động viên bạn, mình tick cho. Cảm ơn bạn nhiều