a)Cho A= \(\dfrac{2015}{2016} \dfrac{2016}{2017} \dfrac{2017}{2018} \dfrac{2018}{2019} \dfrac{2019}{2020} \dfrac{2021}{2015}\) Chứng minh A>6 b)Cho C=\(\dfrac{1}{3} \dfrac{1}{3^2} \dfrac{1}{3^3} ......

ST

Sarah Trần

3 tháng 5 2018

a)Cho A= \(\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2021}{2015}\) Chứng minh A>6 b)Cho C=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+....+\dfrac{1}{3^{2010}}\) Chứng minh rằng C<1 Cho D=\(\dfrac{1}{1^2.2^3}+\dfrac{5}{2^2.3^3}+\dfrac{7}{3^2.4^2}+.....+\dfrac{4019}{2009^2.2010^2}\) Chứng minh rằng D<1 mấy bạn giúp mình nha. Mình cần gấp lắm TT^TT ...

Đọc tiếp

#Toán lớp 7

1

ST

Sarah Trần

3 tháng 5 2018

mấy bạn ơi câu b) là chứng minh C<\(\dfrac{1}{2}\)nha

Đúng(0)

TN

thư nguỹn:>>

20 tháng 4 2022

giúp mk, please :)

\(\dfrac{\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{2022}}{2017+\dfrac{2016}{6}+\dfrac{2015}{7}+...+\dfrac{1}{2021}}\)

A. \(\dfrac{1}{2020}\)

B. \(\dfrac{1}{2021}\)

C. \(\dfrac{1}{2019}\)

D. \(\dfrac{1}{2022}\)

chọn ra 3 ngừi nhanh nhứt:>>

#Toán lớp 6

4

KN

ka nekk

20 tháng 4 2022

d

Đúng(5)

TN

thư nguỹn:>>

20 tháng 4 2022

....

Đúng(1)

NT

Nguyễn Thế sơn

18 tháng 3 2018

So sánh \(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\) và \(B=2018\)

#Toán lớp 6

1

MS

Mashiro Shiina

18 tháng 3 2018

\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)

Đúng(0)

NN

Ngoc Ngan

12 tháng 3 2017

Giai pt:

\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right).3x=1+\dfrac{2019}{2}+\dfrac{2020}{3}+...+\dfrac{4033}{2016}+\dfrac{4034}{2017}\)

#Toán lớp 9

0

HN

hoàng nguyễn phương thảo

28 tháng 1 2019

Cho biểu thức

A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..........+\dfrac{1}{2019}\)

B = \(\dfrac{2018}{1}+\dfrac{2017}{2}+\dfrac{2016}{3}+.......+\dfrac{1}{2018}\) Chứng tỏ rằng \(\dfrac{B}{A}\)là số nguyên

#Toán lớp 7

1

MC

Mặc Chinh Vũ

29 tháng 1 2019

Theo bài ra, ta có: \(B=\dfrac{2018}{1}+\dfrac{2017}{2}+\dfrac{2016}{3}+...+\dfrac{1}{2018}\)

\(B=\left(\dfrac{2018}{1}+1\right)+\left(\dfrac{2017}{2}+1\right)+\left(\dfrac{2016}{3}+1\right)+...+\left(\dfrac{1}{2018}+1\right)-2018\)

\(B=2019+\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}-2018\)

\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\left(2019-2018\right)\)

\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+1\)

\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\dfrac{2019}{2019}\)

\(B=2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)\)

Khi đó:\(\dfrac{B}{A}=\dfrac{2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}}\)

\(\Rightarrow\dfrac{B}{A}=2019\), là 1 số nguyên.

Vậy \(\dfrac{B}{A}\) là số nguyên.

Đúng(0)

KL

Khánh Linh

15 tháng 8 2017

BT1: Cho A = \(\dfrac{1}{2017}+\dfrac{2}{2017^2}+\dfrac{3}{2017^3}+...+\dfrac{2017}{2017^{2017}}+\dfrac{2018}{2017^{2018}}\)

Chứng minh rằng : A < \(\dfrac{2017}{2016^2}\)

#Toán lớp 6

0

TV

Trọng Vũ

6 tháng 8 2017

Đề bài: So sánh

1, \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}với\) 3

2, \(\dfrac{2017}{2018}+\dfrac{2018}{2019}với\dfrac{2017+2018}{2018+2019}\)

#Toán lớp 6

2

LG

Lê Gia Bảo

6 tháng 8 2017

Ta có : \(\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)

Rõ ràng ta thấy : \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\) (1)

\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\) (2)

Từ (1) và (2), suy ra :

\(\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)

Vậy ......................

~ Học tốt ~

Đúng(0)

LG

Lê Gia Bảo

6 tháng 8 2017

Ta có : \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}=\left(1-\dfrac{1}{2018}\right)+\left(1-\dfrac{1}{2019}\right)+\left(1-\dfrac{1}{2020}\right)\)\(=\left(1+1+1\right)-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)\)

\(=3+\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)< 3\)

Vậy \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)

Đúng(0)

NT

Nguyễn Tất Nhật Nam

29 tháng 6 2021

Tính A/B

A=
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\)

B=

\(\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)

#Toán lớp 6

3

SI

Shiba Inu

29 tháng 6 2021

Ta có :

B = \(\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)

B = \(\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)

B = \(\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+1\)

B = \(2021\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+...+\dfrac{1}{2}\right)\) (1)

Mà A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\) (2)

Từ (1) và (2) \(\Rightarrow\) \(\dfrac{A}{B}=\dfrac{1}{2021}\)

Đúng(3)

NL

Nguyễn Lê Phước Thịnh

29 tháng 6 2021

Ta có: \(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)

\(=\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)

\(=\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+\dfrac{2021}{2021}\)

Suy ra: \(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{2021\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=\dfrac{1}{2021}\)

Đúng(2)

Y

ỵyjfdfj

23 tháng 9 2021

\(\dfrac{x+1}{2019}+\dfrac{x+2}{2018}=\dfrac{x+3}{2017}+\dfrac{x+4}{2016}\)

#Toán lớp 7

3

LL

Lấp La Lấp Lánh

23 tháng 9 2021

\(\Leftrightarrow\left(\dfrac{x+1}{2019}+1\right)+\left(\dfrac{x+2}{2018}+1\right)=\left(\dfrac{x+3}{2017}+1\right)+\left(\dfrac{x+4}{2016}+1\right)\)

\(\Leftrightarrow\dfrac{x+2020}{2019}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{2017}-\dfrac{x+2020}{2016}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\)

\(\Leftrightarrow x=-2020\)(do \(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\ne0\))

Đúng(2)

VT

Vũ tũm tĩm

23 tháng 9 2021

Cộng 1 vào mỗi số hạng là ra

Đúng(0)

DM

Dương Minh Hằng

11 tháng 5 2023

so sánh A và B biết:

A=\(\dfrac{2^{2018}}{2^{2018}+3^{2019}}\)+\(\dfrac{3^{2019}}{3^{2019}+5^{2020}}\)+\(\dfrac{5^{2020}}{5^{2020}+2^{2018}}\)

B=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{2019.2020}\).

#Toán lớp 6

1

NL

Nguyễn Lê Phước Thịnh

11 tháng 5 2023

\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)

\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)

=>B<1

=>A>B

Đúng(1)

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP