Cho A =$\frac{1}{1.101}$ $\frac{1}{2.102}$ $\frac{1}{3.103}$ ...... $\frac{1}{25.125}$ B =$\frac{1}{1.26}$ $\frac{1}{2.27}$ $\frac{1}{3.28}$ ....... \(\frac{1}{100...

$A=\frac{\frac{1}{100}.\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{25}-\frac{1}{125}\right)}{\frac{1}{25}.\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+\frac{1}{3}-\frac{1}{28}+...+\frac{1}{100}-\frac{1}{125}\right)}$

$A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-\frac{1}{28}-...-\frac{1}{125}\right)}$

$A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+...+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-...-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{125}\right)}$

$A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}$

$A=\frac{\left(\frac{1}{100}\right)}{\left(\frac{1}{25}\right)}=\frac{1}{4}$

$B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{125}{42}}{\frac{2000}{43}-\frac{250}{252}-\frac{2000}{257}}$

$B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{6000}{2016}}{\frac{2000}{43}-\frac{2000}{2016}-\frac{2000}{257}}$

$B=\frac{16.\left(\frac{1}{9}-\frac{1}{127}+\frac{1}{2017}\right)}{5.\left(\frac{1}{2017}+\frac{1}{9}-\frac{1}{127}\right)}-\frac{6000.\left(\frac{1}{43}-\frac{1}{257}-\frac{1}{2016}\right)}{2000.\left(\frac{1}{43}-\frac{1}{2016}-\frac{1}{257}\right)}$

$B=\frac{16}{5}-3=\frac{1}{5}$

Đặt $C=\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}$

$C=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2005^2}+\frac{1}{2006^2}+\frac{1}{2007^2}$

$C< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2004.2005}+\frac{1}{2005.2006}+\frac{1}{2006.2007}$

$=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2006}+\frac{1}{2006}-\frac{1}{2007}$

$=\frac{1}{4}-\frac{1}{2017}\left(đpcm\right)$

$C>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2005.2006}+\frac{1}{2006.2007}+\frac{1}{2007.2008}$

$=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2006}-\frac{1}{2007}+\frac{1}{2007}-\frac{1}{2008}$

$=\frac{1}{5}-\frac{1}{2008}\left(đpcm\right)$

Vậy $A>\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}>B$

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DG

Dương Gia Linh

12 tháng 9 2018 - olm

Cho: B = $\frac{1}{101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}$

C =$\frac{1}{26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}$

Tìm thương B : C

#Toán lớp 7

0

TL

Trần Lê Việt Hoàng

24 tháng 5 2017

Cho

$A=\dfrac{1}{1.101}+\dfrac{1}{2.102}+\dfrac{1}{3.103}+...+\dfrac{1}{25.125}$

$B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+\dfrac{1}{3.28}+...+\dfrac{1}{100.125}$

Trong đó A có 25 số hạng,B có 100 số hạng.Tìm thương A:B

#Toán lớp 6

1

LG

Lê Gia Bảo

25 tháng 5 2017

Gọi x là thương A:B cần tìm.Theo đề, ta có:

$\left(\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{100.125}\right)x=\dfrac{1}{1.101}+\dfrac{1}{2.102}+...+\dfrac{1}{25.125}$

Nhân 2 vế cho 100, ta có:

$4\left(\dfrac{25}{1.26}+\dfrac{25}{2.27}+...+\dfrac{25}{100.125}\right)x=\dfrac{100}{1.101}+\dfrac{100}{2.102}+...+\dfrac{100}{25.125}$

$\Rightarrow4\left(1-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{100}-\dfrac{1}{125}\right)x=1-\dfrac{1}{101}+\dfrac{1}{2}-\dfrac{1}{102}+...+\dfrac{1}{25}-\dfrac{1}{125}$

$\Rightarrow4\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{125}\right)\right]x=\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{125}\right)$$\Rightarrow4x=1\Rightarrow x=\dfrac{1}{4}$

Vậy hiệu A:B là:$\dfrac{1}{4}$

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QK

Quỷ Khát Máu

11 tháng 5 2016

Tìm x bít

$\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}.x=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{100.110}$

#Toán lớp 6

5

NT

Nhung Trịnh

13 tháng 5 2016

$A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103+...}+\frac{1}{10.110}$

$A=\frac{1}{100}(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110})$

$A=\frac{1}{100}(\frac{1}{1}-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110})$

$A=\frac{1}{100}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}))$ ok?

$B=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}$

$B=\frac{1}{10}(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110})$

$B=\frac{1}{10}(\frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110})$

$B=\frac{1}{10}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{100})-(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}))$=$\frac{1}{10}((\frac{1}{1}+\frac{1}{2}+...+\frac{1}{10})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}))$

B=10A

A.x=10A suy ra x=10

gõ xong mém xỉu. :)

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HN

Hà Như Thuỷ

11 tháng 5 2016

làm biếng gõ quá

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T

tK_nGáO_nGơ

11 tháng 5 2016 - olm

Tìm x , bíÊt:

$\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}x=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{100.110}$

#Toán lớp 6

0

TT

tran thi duyen

15 tháng 1 2015 - olm

Giải phương trình :$\left(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\right).x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}$

#Toán lớp 8

1

NL

Nguyễn Linh Chi

20 tháng 4 2020

Câu hỏi của Huỳnh Ngọc Cẩm Tú - Toán lớp 6 - Học toán với OnlineMath

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NC

Nguyen Chau Minh

5 tháng 5 2017 - olm

Tìm x:

( $\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}$ ) . x = $\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}$

#Toán lớp 6

0

DT

Duong Thi Nhuong

7 tháng 1 2017

Tìm x :

a) $\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{103.105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}$

b) $\left(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\right)x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}$

#Toán lớp 8

3

CC

Chí Cường

7 tháng 1 2017

$\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{103.105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{103}-\frac{1}{105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{1}{2}.\left(1-\frac{1}{105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{52}{105}.\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{52}{105}x-\frac{52}{105}=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow x=-\frac{3}{11}$

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NN

Ngáo Nu

8 tháng 1 2017

b) Đặt $A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}$

A$=\frac{1}{100}\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{1}{3.103}+...+\frac{100}{10.110}\right)$

A$=\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)$

A$=\frac{1}{100}\left[\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]$Đặt $B=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{10}{100.110}$

$B=\frac{1}{10}\left(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)$

$B=\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)$

$B=\frac{1}{10}\left[\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\right]$$=\frac{1}{10}\left[\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]$$B=10A$

$A.x=10A$

$=>x=10$

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TD

Trần Đặng Phan Vũ

14 tháng 3 2018 - olm

$E=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+.........+\frac{1}{10.110}$

$F=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+.........+\frac{1}{100.110}$

tính tỉ số $\frac{E}{F}$

#Toán lớp 6

1

NH

Nguyễn Hưng Phát

14 tháng 3 2018

$100E$$=\frac{100}{1.101}+\frac{100}{2.102}+..........+\frac{100}{10.110}$

$=1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+........+\frac{1}{10}-\frac{1}{110}$

$10F=\frac{10}{1.11}+\frac{10}{2.12}+......+\frac{10}{100.110}$

$=1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+......+\frac{1}{100}-\frac{1}{110}$

$=1+\frac{1}{2}+...+\frac{1}{10}+\frac{1}{11}+....+\frac{1}{100}-\frac{1}{11}-\frac{1}{12}-....-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{110}$

$=1+\frac{1}{2}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}$$=100E$

$\Rightarrow10F=100E\Rightarrow\frac{E}{F}=\frac{1}{10}$

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