K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

18 tháng 5 2020

Giúp mình đi

18 tháng 5 2020

Đặt 2/3 ra ngoài  trong ngoặc còn :

1-1/4+1/4-1/7+...-1/97=96/97

Lấy 2/3 nhân với 96/97 sẽ ra đáp án nhé

26 tháng 4 2018

\(\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{94.97}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{94}-\frac{1}{97}\)

\(=1-\frac{1}{97}\)

\(=\frac{96}{97}\)

mà \(\frac{96}{97}< 1\)

\(\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.07}< 1\)

vậy..................

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{91\cdot94}+\frac{3}{94\cdot97}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\)

\(=1-\frac{1}{97}\)

\(=\frac{96}{97}\)

\(\Rightarrow\frac{96}{97}< 1\)

\(\Rightarrow\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)

Vậy \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)

3 tháng 5 2019

Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\)

\(\Leftrightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{97}\)

\(\Leftrightarrow1-\frac{1}{97}=\frac{96}{97}\)

Do \(\frac{96}{97}< 1\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.97}< 1\)

Vậy:.............................<1

20 tháng 8 2023

\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)

⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)

Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.

9 tháng 4 2015

a)\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)

=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{91}-\frac{1}{94}+\frac{1}{94}-\frac{1}{97}\)(giản ước các phân số giống nhau)

=\(\frac{1}{1}-\frac{1}{97}\)

=\(\frac{96}{97}\)

9 tháng 4 2015

a)    gọi \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.11}+...+\frac{2}{94.97}\)

               \(\Rightarrow\frac{3}{2}A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.97}\)

                     \(\frac{3}{2}A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{97}\)(rút gọn các phân số giống nhau)

                      \(\frac{3}{2}A=\frac{1}{1}-\frac{1}{97}\)

                       \(\frac{3}{2}A=\frac{96}{97}\left(1\right)\)

                       từ \(\left(1\right)\Leftrightarrow A=\frac{96}{97}\div\frac{3}{2}=\frac{64}{97}\)

b)\(\left(1-\frac{1}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{2011}\right)\)

    \(=\frac{6}{7}.\frac{7}{8}.\frac{8}{9}......\frac{2010}{2011}\)

 \(=\frac{6.7.8.9.....2010}{7.8.9......2011}\)(rút gọn các số giống nhau)

\(=\frac{6}{2011}\)

3 tháng 5 2020

\(S_1=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{48\cdot49}+\frac{1}{49\cdot50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

\(S_2=\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+....+\frac{1}{94\cdot97}+\frac{1}{97\cdot100}\)

\(3S_2=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+....+\frac{3}{94\cdot97}+\frac{3}{97\cdot100}\)

\(=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{97}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)

=> \(S_2=\frac{6}{25}:3=\frac{2}{25}\)

16 tháng 5 2017

Đặt A = \(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{91.94}+\dfrac{2}{94.97}\)

\(\dfrac{3A}{2}\)= \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{91.94}+\dfrac{3}{94.97}\)

\(\dfrac{3A}{2}=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{91}-\dfrac{1}{94}+\dfrac{1}{94}-\dfrac{1}{97}\)

\(\dfrac{3A}{2}\) = \(\dfrac{1}{1}-\dfrac{1}{97}\)

\(\dfrac{3A}{2}\) = \(\dfrac{96}{97}\)

3A = \(\dfrac{96}{97}.2\)

3A = \(\dfrac{192}{97}\)

A = \(\dfrac{192}{97}:3\)

A = \(\dfrac{64}{97}\)

Vậy A = \(\dfrac{64}{97}\)

16 tháng 5 2017

\(\dfrac{2}{1.4}\)+\(\dfrac{2}{4.7}\)+\(\dfrac{2}{7.10}\)+....+\(\dfrac{2}{91.94}\)

=2.(1/1.4+1/4.7+1/7.10+...+1/91.94)

=2.\(\dfrac{3}{3}\).(1/1.4+1/4.7+1/7.10+...+1/91.94)

=\(\dfrac{2}{3}\).(3/1.4+3/4.7+3/7.10+...+3/91.94)

=\(\dfrac{2}{3}\).(1/1-1/4+1/4-1/7+1/7-1/10+.....1/91-1/94

=\(\dfrac{2}{3}\).(1-1/94)

=\(\dfrac{2}{3}\).93/94

=31/47

5 tháng 4 2015

Gọi biểu thức sau là A, ta có:

A=(5/1.4)+(5/4.7)+(5/7.10)+...+(5/91.94)

2A=(10/1.4)+(10/4.7)+(10/7.10)+...+(10/91.94)

2A=5/1-5/4+5/4-5/7+5/7-5/10+...+5/91-5/94

2A=5/1-5/4+5/4-5/7+5/7-5/10+...+5/91-5/94

2A=5/1-5/94

2A=465/94

=>A=465/94:2

=>A= tự tính nhé

 

5 tháng 4 2015

\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{91.94}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{91.94}\right)\)

\(=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{94}\right)\)

\(=\frac{5}{3}.\left(1-\frac{1}{94}\right)=\frac{5}{3}.\frac{93}{94}=\frac{155}{94}\)

2 tháng 3 2023

`3/1.4+3/4.7+3/7.10+...+3/94.97`

`=1/1-1/4+1/4-1/7+1/7-1/10+...+1/94-1/97`

`=1-1/97`

`=96/97`

2 tháng 3 2023

\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\\ =1-\dfrac{1}{97}=\dfrac{96}{97}\)