K
Khách

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

18 tháng 3 2021

\(\frac{A}{3}=\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{91-91}{91.94}\)

\(\frac{A}{3}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{94}=1-\frac{1}{94}=\frac{93}{94}\Rightarrow A=\frac{3.93}{94}\)

16 tháng 8 2023

\(B=\dfrac{9}{1\cdot4}+\dfrac{9}{4\cdot7}+...+\dfrac{9}{97\cdot100}\)

\(B=3\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{97\cdot100}\right)\)

\(B=3\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(B=3\cdot\left(1-\dfrac{1}{100}\right)\)

\(B=3\cdot\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)

\(B=3\cdot\dfrac{99}{100}\)

\(B=\dfrac{297}{100}\)

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)

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

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

24 tháng 4 2019

\(A=\frac{9}{1.4}+\frac{9}{4.7}+...+\frac{9}{53.56}\)

\(\Rightarrow\frac{1}{3}A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{53.56}\)

\(\Rightarrow\frac{1}{3}A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{53}-\frac{1}{56}\)

\(\Rightarrow\frac{1}{3}A=1-\frac{1}{56}\)

\(\Rightarrow\frac{1}{3}A=\frac{55}{56}\)

\(\Rightarrow A=\frac{55}{56}\times3\)

\(\Rightarrow A=\frac{165}{56}\)

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

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

4 tháng 8 2015

\(A=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{91\cdot94}=\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{91\cdot94}\right)\)

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

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

\(=\frac{1}{3}\left[\left(\frac{94}{94}-\frac{1}{94}\right)+0+...+0\right]=\frac{1}{3}\cdot\frac{93}{94}=\frac{93}{282}\)

20 tháng 4 2018

\(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + ... + \(\dfrac{3}{91.94}\)

= 1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{91}\) - \(\dfrac{1}{94}\)

= 1 - \(\dfrac{1}{94}\)

= \(\dfrac{93}{94}\)

20 tháng 4 2018

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

= \(3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+....+\dfrac{1}{94}-\dfrac{1}{97}\right)\)

= \(3.\left(1-\dfrac{1}{97}\right)\)

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

= \(\dfrac{288}{97}\)

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.