K
Khách

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

21 tháng 2 2018

Trả lời hô mình

25 tháng 2 2022

\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}\\ =\dfrac{49}{100}\)

25 tháng 2 2022

\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+....+\dfrac{2}{98.100}\)\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{98}-\dfrac{1}{100}\)

                                                   \(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{50}-\dfrac{1}{52}=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{25}{52}\)

25 tháng 7 2023

Bài 1 :

\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)

Ta thấy :

\(3=2^2-1\)

\(15=4^2-1\)

\(35=6^2-1\)

.....

\(9999=100^2-1\)

\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)

\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)

\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)

25 tháng 7 2023

nhanh len nhé mik đang cần gấp ai lam trước mik tích cho

 

17 tháng 8 2016

\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2004.2006}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2004}-\frac{1}{2006}\)

\(=\frac{1}{2}-\frac{1}{2006}\)

\(=\frac{1003}{2006}-\frac{1}{2006}\)

\(=\frac{1002}{2006}\)

\(=\frac{501}{1003}\)

15 tháng 9 2023

a. \(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) + \(\dfrac{1}{6.8}\) + ...... + \(\dfrac{1}{20.22}\)

= 1/2 ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ..... + 1/20 - 1/22)

=1/2 ( 1/2 - 1/22)

= 1/2 . 5/11

= 5/22

b. 1+ 2/3 + 2/6 + 2/10 +...+ 2/45

=>1/2.(1+2/3+2/6+....+2/45)=1/2+2/6+2/12+...+2/90

=1/2+2/2.3+2/3.4+...+2/9.10
=2.(1/4+3-2/2.3+4-3/3.4+...+10-9/9.10)

=2. ( 1/4+1/2-1/3+1/3-1/4+.....+1/9-1/10)
= 2.( 1/4-1/10)=2.3/20=3/10

=> vì 1/2.*=3/10

=> *=3/10:1/2=3/5

tick mình nhé

 

16 tháng 9 2023

B = 1 + \(\dfrac{2}{3}\) + \(\dfrac{2}{6}\) +\(\dfrac{2}{10}\) + \(\dfrac{2}{15}\)+...+ \(\dfrac{2}{45}\)

B = 1 + 2.(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\)+...+ \(\dfrac{1}{45}\))

B = 1 + \(\dfrac{4}{2}\).(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) +\(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ...+ \(\dfrac{1}{45}\))

B = 1 + 4.( \(\dfrac{1}{6}\) +\(\dfrac{1}{12}\)\(\dfrac{1}{20}\)\(\dfrac{1}{30}\)+...+ \(\dfrac{1}{90}\))

B = 1 + 4.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\)\(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{9.10}\))

B = 1 + 4 .( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\)+...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\))

B = 1 + 4.( \(\dfrac{1}{2}\) - \(\dfrac{1}{10}\))

B = 1 + 4. \(\dfrac{2}{5}\)

B = \(\dfrac{13}{5}\)

 

Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)

Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)

\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)

\(=2\cdot\dfrac{505}{1011}\)

\(=\dfrac{1010}{1011}\)

a) Ta có: \(\dfrac{1}{2022}-\dfrac{5}{2\cdot4}-\dfrac{5}{4\cdot6}-\dfrac{5}{6\cdot8}-...-\dfrac{5}{2020\cdot2022}\)

\(=\dfrac{1}{2022}-5\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2020\cdot2022}\right)\)

\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2020\cdot2022}\right)\)

\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)

\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)

\(=\dfrac{1}{2022}-\dfrac{5}{2}\cdot\dfrac{1010}{2022}\)

\(=\dfrac{1}{2022}-\dfrac{2025}{2022}=\dfrac{-1262}{1011}\)

b) Ta có: \(\dfrac{2^2}{1\cdot3}+\dfrac{2^2}{3\cdot5}+...+\dfrac{2^2}{197\cdot199}\)

\(=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{197\cdot199}\right)\)

\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{197}-\dfrac{1}{199}\right)\)

\(=2\left(1-\dfrac{1}{199}\right)\)

\(=2\cdot\dfrac{198}{199}=\dfrac{396}{199}\)