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29 tháng 3 2017

\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{17\cdot21}< 1\)

\(A=\dfrac{4}{4}\cdot\left(\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+...+\dfrac{1}{17\cdot21}\right)< 1\)

\(A=\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{17}-\dfrac{1}{21}< 1\)

\(A=1-\dfrac{1}{21}< 1\) (đúng) (đpcm).

29 tháng 3 2017

Đề sai

29 tháng 3 2017

Tìm x

\(\dfrac{x}{5}\)=\(\dfrac{x+6}{15}\)

\(\Rightarrow\)\(\dfrac{3x}{15}\)=\(\dfrac{x+6}{15}\)

\(\Rightarrow\)3x = x+6

\(\Rightarrow\)2x=6

\(\Rightarrow\)x=3

TÍNH TỔNG S

S=\(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{17.21}\)

S=\(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{17}-\dfrac{1}{21}\)

S= \(1-\dfrac{1}{21}\)

S= \(\dfrac{20}{21}\)

29 tháng 3 2017

Tìm x:

\(\dfrac{x}{5}=\dfrac{x+6}{15}=>\dfrac{3x}{15}=\dfrac{x+6}{15}\)

=> 3x = 6 + x

=> 2x = 6

=> x = 3

Tính tổng S:

\(S=\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{17.21}\)

\(S=\dfrac{4}{1}-\dfrac{4}{5}+\dfrac{4}{5}-\dfrac{4}{9}+\dfrac{4}{9}-\dfrac{4}{13}+...+\dfrac{4}{17}-\dfrac{4}{21}\)

\(S=4-\dfrac{4}{21}\)

\(S=\dfrac{80}{21}\)

20 tháng 2 2023

Tính chất của phân số bạn cần biết như sau:

\(\dfrac{b-a}{a\cdot b}=\dfrac{1}{a}-\dfrac{1}{b}\)

Gọi biểu thức trên là A ,ta có:

\(A=\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+\dfrac{1}{13\cdot17}+\dfrac{1}{17\cdot21}+\dfrac{1}{21\cdot25}\)

\(4A=\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+\dfrac{4}{13\cdot17}+\dfrac{4}{17\cdot21}+\dfrac{4}{21\cdot25}\)

\(4A=\dfrac{9-5}{5\cdot9}+\dfrac{13-9}{9-13}+\dfrac{17-13}{13\cdot17}+\dfrac{21-17}{17\cdot21}+\dfrac{25-21}{21\cdot25}\)

Áp dụng tính chất phân số đã nêu ở trên, ta được:

\(4A=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{25}\)

\(4A=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{5}{25}-\dfrac{1}{25}=\dfrac{4}{25}\)

\(A=4A:4=\dfrac{4}{25}:4=\dfrac{16}{25}\)

Vậy \(A=\dfrac{16}{25}\)

25 tháng 2 2023

\(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{21.25}\\ =\dfrac{4\cdot\dfrac{1}{4}}{5.9}+\dfrac{4\cdot\dfrac{1}{4}}{9.13}+...+\dfrac{4\cdot\dfrac{1}{4}}{21.25}\\ =\dfrac{1}{4}\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{21.25}\right)\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{21}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)=\dfrac{1}{4}\left(\dfrac{5}{25}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\cdot\dfrac{4}{25}=\dfrac{1}{25}\)

25 tháng 2 2023

`1/(5.9) + 1/(9.13) + ...+ 1/(21.25)`

`= 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/21 - 1/25`

`= 1/5 - 1/25`

`= 4/25`

 

25 tháng 3 2017

\(x+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{41.45}=-\dfrac{37}{45}\\ x+\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{44}-\dfrac{1}{45}\right)=-\dfrac{37}{45}\\ x+\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=-\dfrac{37}{45}\\ x+\dfrac{8}{45}=-\dfrac{37}{45}\\ x=-\dfrac{37}{45}-\dfrac{8}{45}\\ x=-1\)

30 tháng 3 2017

bạn học nhà cô Phương à

a: \(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{121}-\dfrac{1}{124}=1-\dfrac{1}{124}=\dfrac{123}{124}\)

b: \(=3\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)=3\cdot\dfrac{99}{202}=\dfrac{297}{202}\)

c: \(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{401}-\dfrac{1}{405}\right)=\dfrac{1}{4}\cdot\dfrac{404}{405}=\dfrac{101}{405}\)

d: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)

1 tháng 3 2022

đề bài là j

AH
Akai Haruma
Giáo viên
12 tháng 5 2021

Lời giải:

\(2A=\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+...+\frac{1}{29}-\frac{1}{41}=1-\frac{1}{41}=\frac{40}{41}\)

\(\Rightarrow A=\frac{20}{21}\)

\(3B=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\)

\(=1-\frac{1}{31}=\frac{30}{31}\)

\(\Rightarrow B=\frac{10}{31}=\frac{20}{62}<\frac{20}{41}\)

Do đó $A>B$

13 tháng 5 2021

C ưi , c hỗ trợ câu em mới gửi vào inb nhé

11 tháng 5 2021

A.2=4/1.5+6/5.11+...+12/29.41

A.2=1-1/5+1/5-1/11+...+1/29-1/41

A.2=1-1/41

A.2=40/41

A=20/41

B.3=3/1.4+6/4.10+...+12/29.31

B.3=1-1/4+1/4-1/10+...+1/29-1/31

B.3=1-1/31

B.3=30/31

B=10/31

Vì 20/41.10/31 nên A>B

11 tháng 5 2021

\(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)

\(\Rightarrow2A=\dfrac{4}{1.5}+\dfrac{6}{5.11}+\dfrac{8}{11.19}+\dfrac{10}{19.29}+\dfrac{12}{29.41}\)

\(\Rightarrow2A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{41}\)

\(\Rightarrow2A=1-\dfrac{1}{41}=\dfrac{40}{41}\)

\(\Rightarrow A=\dfrac{40}{41}:2=\dfrac{20}{41}\)(1)

\(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)

\(\Rightarrow3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\)

\(\Rightarrow3B=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}\)

\(\Rightarrow3B=\dfrac{1}{1}-\dfrac{1}{31}=\dfrac{30}{31}\)

\(\Rightarrow B=\dfrac{30}{31}:3=\dfrac{10}{31}\)

\(\Rightarrow B=\dfrac{2}{2}.\dfrac{10}{31}=\dfrac{20}{62}\)

+)Ta có:\(\dfrac{20}{62}< \dfrac{20}{41}\Rightarrow B< A\)

Hay A>B(ĐPCM)

Chúc bn học tốt

21 tháng 4 2020

Chứng minh \(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}< 1\)

\(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)

\(A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)

\(A=\frac{1}{1}-\frac{1}{21}\)

\(A=\frac{20}{21}\)

\(\frac{20}{21}< 1\)

=> \(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}< 1\)( đpcm ) 

* Mình sợ sai xD *

$\dfrac{4}{1.4}+\dfrac{5.9}+....+\dfrac{4}{2001.2005}$
$=1+\dfrac15-\dfrac19+....+\dfrac{1}{2001}-\dfrac{1}{2005}$
$=1-\dfrac{1}{2005}=\dfrac{2004}{2005}$

6 tháng 7 2021

 \(\dfrac{4}{1.4}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)

\(=1+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\)

\(=1+\dfrac{1}{5}-\dfrac{1}{2005}\)

\(=1+\dfrac{401}{2005}-\dfrac{1}{2005}\)

\(=1+\dfrac{400}{2005}=1+\dfrac{80}{401}=\dfrac{481}{401}\)