A=\(\dfrac{2008^{2008}+1}{2008^{2009}+1}\) B=\(\dfrac{2008^{2007}+1}{2008^{2008}+1}\)

K

Khách

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PT

Phạm Thanh Vân

23 tháng 12 2022

A=\(\dfrac{2008^{2008}+1}{2008^{2009}+1}\) B=\(\dfrac{2008^{2007}+1}{2008^{2008}+1}\)

2

DD

Dương đình minh

26 tháng 12 2022

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DD

Dương đình minh

26 tháng 12 2022

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Các câu hỏi dưới đây có thể giống với câu hỏi trên

YA

Yui Arayaki

13 tháng 2 2018

Tính tỉ số \(\dfrac{A}{B}\) , biết:

\(A=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\)

\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)

1

TM

Trần Minh Hoàng

29 tháng 11 2020

\(B=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\Rightarrow\frac{A}{B}=\frac{1}{2009}\)

ND

Natsu Dragneel

26 tháng 2 2018

Tính tỉ số A/B, biết :

A\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\)

B\(=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)

1

BN

Bích Ngọc

26 tháng 2 2018

\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\)

\(B=1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)\)

\(B=\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+..+\dfrac{2009}{2007}+\dfrac{2009}{2008}\)

\(B=2009\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)

\(\dfrac{A}{B}=\dfrac{1}{2009}\)

B

Black

20 tháng 12 2017

1)Tính tỉ số \(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)

2) chứng minh: \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+..+\dfrac{19}{9^2.10^2}< 1\)

1

MS

Mashiro Shiina

20 tháng 12 2017

1)\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2008+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)}\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2007}+\dfrac{2009}{2008}}\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)

\(\dfrac{A}{B}=\dfrac{1}{2009}\)

2) \(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)

\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

\(A=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)

CN

Cô Nàng Họ Dương

7 tháng 7 2016 - olm

Bài 1: So sánh A và B

\(A=\frac{2008^{2008}+1}{2008^{2009}+1}\)

\(B=\frac{2008^{2007}+1}{2008^{2008}+1}\)

0

NM

Nguyễn Minh Dương

VIP

21 tháng 9 2023

So sánh bt: \(A=\dfrac{2008^{2008}+1}{2008^{2009}+1};B=\dfrac{2008^{2007}+1}{2008^{2008}+1}\)

1

T

Toru

21 tháng 9 2023

\(A=\dfrac{2008^{2008}+1}{2008^{2009}+1}\)

\(2008\cdot A=\dfrac{2008^{2009}+2008}{2008^{2009}+1}\)

\(=\dfrac{2008^{2009}+1+2007}{2008^{2009}+1}\)

\(=1+\dfrac{2007}{2008^{2009}+1}\)

\(B=\dfrac{2008^{2007}+1}{2008^{2008}+1}\)

\(2008\cdot B=\dfrac{2008^{2008}+2008}{2008^{2008}+1}\)

\(=\dfrac{2008^{2008}+1+2007}{2008^{2008}+1}\)

\(=1+\dfrac{2007}{2008^{2008}+1}\)

Ta có: \(2008^{2009}+1>2008^{2008}+1\)

\(\Rightarrow\dfrac{1}{2008^{2009}+1}< \dfrac{1}{2008^{2008}+1}\)

\(\Rightarrow\dfrac{2007}{2008^{2009}+1}< \dfrac{2007}{2008^{2008}+1}\)

\(\Rightarrow1+\dfrac{2007}{2008^{2009}+1}< 1+\dfrac{2007}{2008^{2008}+1}\)

hay \(A < B\)

#\(Toru\)

DO

꧁Dazai ~ Osamu꧂

20 tháng 12 2019 - olm

So sánh A vs B

\(A=\frac{2008^{2008}+1}{2008^{2009}+1}\)

\(B=\frac{2008^{2007}+1}{2008^{2008}+1}\)

2

NH

Nhật Hạ

20 tháng 12 2019

ý, nếu không được dùng cách kia thì làm cách này cho chắc đi :v

Ta có: \(2008A=\frac{2008\left(2008^{2008}+1\right)}{2008^{2009}+1}=\frac{2008^{2009}+2008}{2008^{2009}+1}=\frac{\left(2008^{2009}+1\right)+2007}{2008^{2009}+1}=1+\frac{2007}{2008^{2009}+1}\)

Lại có: \(2008B=\frac{2008\left(2008^{2007}+1\right)}{2008^{2008}+1}=\frac{2008^{2008}+2008}{2008^{2008}+1}=\frac{\left(2008^{2008}+1\right)+2007}{2008^{2008}+1}=1+\frac{2007}{2008^{2008}+1}\)

Vì 2008 < 2009 \(\Rightarrow2008^{2008}< 2008^{2009}\)\(\Rightarrow2008^{2008}+1< 2008^{2009}+1\)\(\Rightarrow\frac{2007}{2008^{2008}+1}>\frac{2007}{2008^{2009}+1}\)\(\Rightarrow1+\frac{2007}{2008^{2008}+1}>1+\frac{2007}{2008^{2009}+1}\)\(\Rightarrow2008B>2008A\)\(\Rightarrow B>A\)

NH

Nhật Hạ

20 tháng 12 2019

Vì A <1 , B < 1

Nên ta có: \(A=\frac{2008^{2008}+1}{2008^{2009}+1}< \frac{2008^{2008}+1+2007}{2008^{2009}+1+2007}=\frac{2008^{2008}+2008}{2008^{2009}+2008}=\frac{2008\left(2008^{2007}+1\right)}{2008\left(2008^{2008}+1\right)}=\frac{2008^{2007}+1}{2008^{2008}+1}=B\)

TC

Thầy Cao Đô

Giáo viên VIP

22 tháng 12 2022 - olm

Câu hỏi hay

Câu 19. Cho ${A}=\dfrac{2008^{2008}+1}{2008^{2009}+1}$; ${B}=\dfrac{2008^{2007}+1}{2008^{2008}+1}$. So sánh $A$ và $B$.

5

DT

Đoàn Trần Quỳnh Hương

22 tháng 12 2022

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VT

Vũ Tiến Dũng

23 tháng 12 2022

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NT

Nguyễn Thị Thanh Phương

24 tháng 9 2016

So Sánh

a,A= \(\frac{2008^{2008}+1}{2008^{2009}+1}\)và B=\(\frac{2008^{2007}+1}{2008^{2008}+1}\)

b, M=\(\frac{100^{100}+1}{100^{99}+1}\)và N= \(\frac{100^{101}+1}{100^{100}+1}\)

1

SG

soyeon_Tiểubàng giải

24 tháng 9 2016

a) Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\) (a;b;m \(\in\) N*)

Ta có:

\(A=\frac{2008^{2008}+1}{2008^{2009}+1}< \frac{2008^{2008}+1+2007}{2009^{2009}+1+2007}\)

\(A< \frac{2008^{2008}+2008}{2008^{2009}+2008}\)

\(A< \frac{2008.\left(2008^{2007}+1\right)}{2008.\left(2008^{2008}+1\right)}=\frac{2008^{2007}+1}{2008^{2008}+1}=B\)

=> A < B

b) Áp dụng \(\frac{a}{b}>1\Leftrightarrow\frac{a}{b}>\frac{a+m}{b+m}\) (a;b;m \(\in\) N*)

Ta có:

\(N=\frac{100^{101}+1}{100^{100}+1}>\frac{100^{101}+1+99}{100^{100}+1+99}\)

\(N>\frac{100^{101}+100}{100^{100}+100}\)

\(N>\frac{100.\left(100^{100}+1\right)}{100.\left(100^{99}+1\right)}=\frac{100^{100}+1}{100^{99}+1}=M\)

=> M > N

NT

Nguyễn Thị Thanh Phương

24 tháng 9 2016

Cảm ơn bạn nhiều

TK

Thuy Khuat

22 tháng 11 2017

Cho A=\(\dfrac{2006}{2007}+\dfrac{2007}{2008}+\dfrac{2008}{2009}+\dfrac{2009}{2006}\) .Hãy so sánh số đó với 4

1

MC

Mặc Chinh Vũ

6 tháng 3 2019

\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)

\(A=\left(1-\frac{1}{2007}\right)+\left(1-\frac{1}{2008}\right)+\left(1-\frac{1}{2009}\right)+\left(1+\frac{3}{2006}\right)\)

\(A=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}+1+\frac{3}{2006}\)

\(A=\left(1+1+1+1\right)-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)

\(A=4-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)

Ta có: \(\left\{{}\begin{matrix}\frac{1}{2007}< \frac{1}{2006}\\\frac{1}{2008}< \frac{1}{2006}\\\frac{1}{2009}< \frac{1}{2006}\end{matrix}\right.\Rightarrow\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}< \frac{1}{2006}+\frac{1}{2006}+\frac{1}{2006}=\frac{3}{2006}\)

\(\Rightarrow\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}< 0\)

\(\Rightarrow4-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)>4\)

hay \(A>4\)

\(\text{Vậy A>4}\)