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

6 tháng 7 2021

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

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

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

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

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

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

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

27 tháng 3 2017

Theo quy luật thì mình nghĩ đáng lẽ \(\dfrac{4}{5.9}\)phải là\(\dfrac{4}{7.9}\)Bạn có chép sai đề ko?

27 tháng 3 2017

A=1-\(\dfrac{4}{5.7}-\dfrac{4}{7.9}-\dfrac{4}{9.11}...-\dfrac{4}{59.61}\)

A=\(1-\left(\dfrac{4}{5.7}+\dfrac{4}{7.9}+\dfrac{4}{9.11}+...+\dfrac{4}{59.61}\right)\)

Đặt B=\(\dfrac{4}{5.7}+\dfrac{4}{7.9}+\dfrac{4}{9.11}+...+\dfrac{4}{59.61}\)

B=\(2\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{59.61}\right)\) B=\(2\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\) B=\(2\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=2.\dfrac{56}{305}\) B=\(\dfrac{112}{305}\) \(\Rightarrow A=1-\dfrac{112}{305}=\dfrac{193}{305}\)

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

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

\(\dfrac{4}{3.5}+\dfrac{8}{5.9}+\dfrac{12}{9.15}+...+\dfrac{32}{x\left(x+16\right)}=\dfrac{16}{15}\)

\(2.\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+\dfrac{6}{9.15}+..+\dfrac{16}{X.\left(X+16\right)}\right)=\dfrac{16}{15}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{X}-\dfrac{1}{X+16}=\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{1}{3}-\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{-1}{5}\)

\(X+16=-5\)

\(X=-21\)

 

10 tháng 5 2023

\(\dfrac{3}{1\cdot4}+\dfrac{4}{4\cdot8}+\dfrac{5}{8\cdot13}+..........+\dfrac{10}{43\cdot53}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{13}+....................+\dfrac{1}{43}-\dfrac{1}{53}\)

\(=1-\dfrac{1}{53}\)

\(=\dfrac{52}{53}\)

 

6 tháng 7 2015

S=4/1.5+4/5.9+...+4/2001.2005

S =1/1 - 1/5 + 1/5 -1/9 + ...+ 1/2001 - 1/2005

S = 1/1 - 1/2005 

S = 2014/2015