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9 tháng 4 2017

a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)

=\(1-\frac{1}{56}=\frac{55}{56}\)

b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)

\(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)

=> \(\frac{1}{3}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)

=> \(A=\frac{99}{100}.3=\frac{297}{100}\)

c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)

=\(1-\frac{1}{103}=\frac{102}{103}\)

d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)

=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)

=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)

=\(1-\frac{1}{103}=\frac{102}{103}\)

=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)

e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)

=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)

=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)

=\(1-\frac{1}{105}=\frac{104}{105}\)

=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)

9 tháng 4 2017

a) \(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...\dfrac{10}{46.56}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{46}-\dfrac{1}{56}\)

\(P=1-\dfrac{1}{56}\)

\(P=\dfrac{55}{56}\)

b) \(A=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{99.100}\)

\(A=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(A=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

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

\(A=3.\dfrac{99}{100}\)

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

c) \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)

\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(B=1-\dfrac{1}{103}\)

\(B=\dfrac{102}{103}\)

d) \(C=\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{100.103}\)

\(C=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\right)\)

\(C=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(C=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)

\(C=\dfrac{5}{3}.\dfrac{102}{103}\)

\(C=\dfrac{170}{103}\)

e) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)

\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}.\dfrac{104}{105}\)

\(D=\dfrac{26}{15}\)

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

19 tháng 3

a; \(\dfrac{-1}{n}\) - \(\dfrac{1}{n+a}\) 

\(\dfrac{-n-a-n}{n.\left(n+a\right)}\)

\(\dfrac{-2n-a}{n.\left(n+a\right)}\)

b; \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ....+ \(\dfrac{1}{2007.2008}\)

\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)

\(\dfrac{1}{1}\) - \(\dfrac{1}{2008}\)

\(\dfrac{2007}{2008}\)

c; \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)

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

\(\dfrac{1}{1}\) - \(\dfrac{1}{97}\)

\(\dfrac{96}{97}\)

24 tháng 3 2019

a, \(\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{99.100}\)

=9.(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\))

= 9(1 -\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\))

=9(1-\(\frac{1}{100}\))

A=\(\frac{891}{100}\)

b, \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)

=1-(\(\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{30}\))

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

B=\(\frac{29}{30}\)

24 tháng 3 2019

a) \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+...+\dfrac{9}{99.100}\)

\(=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=9\left(1-\dfrac{1}{100}\right)\)

\(=9.\dfrac{99}{100}\)

\(=\dfrac{891}{100}\)

b) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{27.30}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{27}-\dfrac{1}{30}\)

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

\(=\dfrac{29}{30}\)

15 tháng 6 2018

Giải:

b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}\)

\(=\dfrac{1}{1}-\dfrac{1}{2009}\)

\(=\dfrac{2008}{2009}\)

c) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{4}{7.10}+...+\dfrac{3}{94.97}\)

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

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

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

Vậy ...

Các câu sau tương tự

16 tháng 6 2018

b, \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2008.1009}\)

\(=\)\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2008}-\dfrac{1}{2009}\)

\(=\dfrac{1}{1}-\dfrac{1}{2009}=\dfrac{2009}{2009}-\dfrac{1}{2009}=\dfrac{2008}{2009}\)

8 tháng 8 2018

Làm từng phần nha bạn

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{298\cdot301}+x=\frac{299}{301}\)

Đặt \(A+x=\frac{299}{301}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{298}-\frac{1}{301}\)

\(A=1-\frac{1}{301}\)

\(A=\frac{300}{301}\)

=> \(\frac{300}{301}+x=\frac{299}{301}\)

\(x=\frac{299-300}{301}\)

\(x=-\frac{1}{301}\)

8 tháng 8 2018

\(A=5\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{301\cdot304}\right)\)

\(\frac{3A}{5}=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{301\cdot304}\)

\(\frac{3}{5}\cdot A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{301}-\frac{1}{304}\)

\(\frac{3}{5}\cdot A=1-\frac{1}{304}\)

\(\frac{3}{5}\cdot A=\frac{303}{304}\)

\(A=\frac{505}{304}\)

15 tháng 8 2023

a/

3A=1.2.3+2.3.3+3.4.3+...+98.99.3=

=1.2.3+2.3.(4-1)+3.4.(5-2)+...+98.99.(100-97)=

=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-97.98.99+98.99.100=

=98.99.100=> A=98.33.100

b

6B=1.3.6+3.5.6+5.7.6+...+99.101.6=

=1.3.(5+1)+3.5.(7-1)+5.7.(9-3)+...+99.101.(103-97)=

=1.3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-97.99.101+99.101.103=

=1.3+99.101.103=> (3+99.101.103):6

c/

9S=1.4.9+4.7.9+7.10.9+...+2017.2020.9=

=1.4.(7+2)+4.7.(10-1)+7.10.(13-4)+...+2017.2020.(2023-2014)=

=1.2.4+1.4.7-1.4.7+4.7.10--4.7.10+7.10.13-...-2014.2017.2020+2017.2020.2023=

=1.2.4+2017.2020.2023=> S=(2.4+2017.2020.2023):9

Dạng tổng quát: tính tổng các tích có quy luật: các thừa số của các tích lập thành dãy số cách đều. các thừa số đầu tiên của số hạng liền sau cũng chính là các thừa số sau cùng của số hạng liền trước thì ta nhân tổng với số k

Số k được tính theo quy luật \(k=\left(n+1\right)xd\)

            Trong đó: n: số thừa số của 1 số hạng

                            d: Khoảng cách giữa hai thừa số liền kề trong mỗi số hạng

Chúc em học tốt

 

 

14 tháng 7 2015

a)=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\)

\(=1-\frac{1}{2009}\)

\(=\frac{2008}{2009}\)

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

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

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

14 tháng 7 2015

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2008.2009}\) \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2008}-\frac{1}{2009}\)  

= 1 - 1/2009 

= 2008/2009

b) 3/1.4 + 3/4.7 + 3/7.10 + .... + 3/94.97

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

= 1 - 1/97

= 96/97