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23 tháng 3 2018

mình làm được nhưng đánh lâu lắm

8 tháng 1 2017

\(A=xemlai\) chưa hưa hiểu Quy luật

\(B=\frac{\left(n.\left(n+2\right)+1\right)}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)

\(B=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.5}...\frac{98.98}{97.99}\frac{99.99}{98.100}\frac{100.100}{99.101}\\\)

\(B=\frac{2.100}{1.101}=\frac{200}{101}\)

7 tháng 4 2017

Lâm đi là: 35 phút +2 giờ 20phút =2 giờ 55 phút

7 tháng 4 2017

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(A=1-\frac{1}{2017}\)

\(A=\frac{2016}{2017}\)

22 tháng 3 2018

\(Q=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)

\(Q=\left(\frac{1}{2}\right).\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)...\left(\frac{99}{100}\right)\)

\(Q=\frac{1}{100}\)

\(P=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)

\(P=\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)\left(\frac{3.5}{3.5}+\frac{1}{3.5}\right)...\left(\frac{99.101}{99.101}+\frac{1}{99.101}\right)\)

\(P=\left(\frac{4}{1.3}\right)\left(\frac{9}{2.4}\right)\left(\frac{16}{3.5}\right)...\left(\frac{10000}{99.101}\right)\)

\(P=\left(\frac{2^2}{1.3}\right)\left(\frac{3^2}{2.4}\right)\left(\frac{4^2}{3.5}\right)...\left(\frac{100^2}{99.101}\right)\)

Bạn tự tách ra rồi bạn sẽ ra kết quả như ở dưới

\(P=\frac{201}{100}\)

24 tháng 8 2019

Bài làm

D=ko viết lại đề

=1/1.3+1/1.5+1/5.7+1/7.9-1/2.4-1/4.6-1/6.8-1/8.10

=1+1/9-1-1/10

=10/9-9/10

=19/90

=(1/1.3+...+1/7.9)-(1/2.4+...+1/8.10)

=2(1/1.3+...+1/7.9)-2(1/2.4+...+1/8.10)

=(2/1.3+...+2/7.9)-(2/2.4+...+2/8.10)

=(1-1/3+...+1/7-1/9)-(1/2-1/4+   +1/8-1/10)

=1-1/9-1/2+1/10

tự tính tiếp nhé

9 tháng 5 2017

\(S=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{7.9}+\frac{1}{8.10}\)

\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\right)\)

Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\)

2A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{7.9}\)

2A = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\)

2A = \(1-\frac{1}{9}=\frac{8}{9}\)

A = \(\frac{8}{9}:2=\frac{4}{9}\)

Đặt B = \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\)

2B = \(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{8.10}\)

2B = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)

2B = \(\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

B = \(\frac{2}{5}:2=\frac{1}{5}\)

Thay A và B vào S ta được:

\(S=\frac{4}{9}+\frac{1}{5}=\frac{29}{45}\)

9 tháng 5 2017

\(S=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{7.9}+\frac{1}{8.10}\)

\(\Rightarrow S=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}\)

\(S=\frac{1}{2}\left(\frac{8}{9}+\frac{2}{5}\right)\)

\(S=\frac{1}{2}.\frac{58}{45}\)

\(S=\frac{29}{45}\)

8 tháng 9 2016

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

  \(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)

  \(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

\(=\frac{4}{9}-\frac{1}{5}\)

\(=\frac{11}{45}\)

 

8 tháng 9 2016

Cảm ơn giúp  bài nữa nha !!

7 tháng 9 2016

\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)

7 tháng 9 2016

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

\(S=\frac{4}{9}-\frac{1}{5}\)

\(S=\frac{11}{45}\)

24 tháng 10 2016

=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\right)\)

\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right)\)

\(\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)

=\(\frac{29}{45}\)

24 tháng 10 2016

29/45 bạn nhé