CMR: \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}.........+\dfrac{1}{100^2}< \dfrac{1}{4}\)

K

Khách

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KN

Kha Nguyễn

28 tháng 3 2018

CMR:

\(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}.........+\dfrac{1}{100^2}< \dfrac{1}{4}\)

1

KN

Kha Nguyễn

28 tháng 3 2018

https://i.imgur.com/0DG8QJy.png

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

CB

Cherry Bùi

25 tháng 1 2018

CMR: \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)

1

MS

Mashiro Shiina

26 tháng 1 2018

Đặt: \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\)

Ta có: \(\left\{{}\begin{matrix}A>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{6}\\A< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\end{matrix}\right.\)

Vậy \(\dfrac{1}{6}< A< \dfrac{1}{4}\)

MD

Monkey D Luffy

18 tháng 2 2018

Chứng minh rằng : \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+....+\dfrac{1}{100^2}< \dfrac{1}{4}\)

0

OO

O O O

26 tháng 12 2017

Chứng minh rằng : \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)

1

NT

Ngô Tấn Đạt

26 tháng 12 2017

\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\\ =\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}\\ < \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)

\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}\\=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ =\dfrac{1}{5}-\dfrac{1}{101}\)

MS

Mashiro Shiina

26 tháng 12 2017

and.....

FA

Forever alone

4 tháng 4 2017

Chứng minh:

\(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+........+\dfrac{1}{100^2}< \dfrac{1}{4}\)

1

NT

Nguyễn Thị Huyền Trang

6 tháng 4 2017

Đại số lớp 7

TB

Trần Bảo Hân

5 tháng 9 2018

Chứng minh rằng:

\(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)

1

NC

Nguyễn Chơn Nhân

21 tháng 9 2018

$1/6<$ 1/5^2+1/6^2+1/7^2+...+1/100 $^2<1/$ $4$

ta có:

(+)1/5^2+1/6^2+1/7^2+...+1/100^2<1/4.5+1/5.6+...+1/99.100
=1/4-1/5+1/5-...+1/99-1/100

=1/4-1/100<1/4

=>1/5^2+1/6^2+1/7^2+...+1/100^2<1/4

(+)1/5^2+1/6^2+1/7^2+...+1/100^2>1/5.6+...+1/99.100

=1/5-1/6+1/6-...+1/99-1/100

=1/5-1/100>1/6

=>1/5^2+1/6^2+1/7^2+...+1/100^2

NT

Ngô Thành Chung

10 tháng 2 2018

Chứng minh rằng

\(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)

1

NT

Ngô Tấn Đạt

10 tháng 2 2018

\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+....+\dfrac{1}{100^2}\\ >\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}\\ =\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ =\dfrac{1}{5}-\dfrac{1}{101}\\ =\dfrac{96}{505}\\ >\dfrac{1}{6}\)

\(\dfrac{1}{5^2}+...+\dfrac{1}{100^2}\\ < \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+....+\dfrac{1}{99.100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)

TN

Thạch Nguyễn

5 tháng 7 2017

CMR \(\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+......+\dfrac{2}{100^2}< \dfrac{1}{2}\)

4

NT

Nguyễn Thị Huyền Trang

5 tháng 7 2017

Ta có: \(\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+...+\dfrac{1}{100^2}< \dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{99.100}\)

\(=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\right)\)

\(=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2.\left(\dfrac{1}{4}-\dfrac{1}{100}\right)< 2.\dfrac{1}{4}=\dfrac{1}{2}\)

hay \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)

Vậy ...

MV

Mới vô

5 tháng 7 2017

Gọi \(A=\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+...+\dfrac{2}{100^2}\)

\(A< \dfrac{2}{4\cdot5}+\dfrac{2}{5\cdot6}+\dfrac{2}{6\cdot7}+...+\dfrac{2}{99\cdot100}\\ A< 2\cdot\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{99\cdot100}\right)\\ A< 2\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ A< 2\cdot\left(\dfrac{1}{4}-\dfrac{1}{100}\right)\\ A< \dfrac{1}{2}-\dfrac{1}{50}\\ A< \dfrac{12}{25}< \dfrac{1}{2}\)

Vậy \(\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+...+\dfrac{2}{100^2}< \dfrac{1}{2}\)

DX

dream XD

2 tháng 7 2021

Tính giá trị biểu thức A , biết rằng A = M : N Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\) N...

1

NL

Nguyễn Lê Phước Thịnh

2 tháng 7 2021

Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

LT

Lê Tuấn Nghĩa

20 tháng 10 2017

Cho B = \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{5}{6}.\)...\(.\dfrac{99}{100}\). CMR \(\dfrac{1}{15}< B< \dfrac{1}{10}\)

0