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9 tháng 6 2015

B=1/51.100+1/52.99+...+1/100.51

=>151B=1/51+1/100+1/52+1/99+...+1/100+1/51

=>151B/2=1/51+1/52+1/53+1/54+...+1/100

=>B=2/151.(1/51+1/52+1/53+1/54+...+1/100)

AH
Akai Haruma
Giáo viên
18 tháng 6 2019

Lời giải:

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}\)

\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Mặt khác:

\(151B=\frac{51+100}{51.100}+\frac{52+99}{52.99}+....+\frac{99+52}{99.52}+\frac{100+51}{100.51}\)

\(=\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+....+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\)

\(=\left(\frac{1}{100}+\frac{1}{99}+....+\frac{1}{52}+\frac{1}{51}\right)+\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)\)

\(=2\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)=2A\)

\(\Rightarrow \frac{A}{B}=\frac{151}{2}\)

AH
Akai Haruma
Giáo viên
20 tháng 6 2019

Lời giải:

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}\)

\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Mặt khác:

\(151B=\frac{51+100}{51.100}+\frac{52+99}{52.99}+....+\frac{99+52}{99.52}+\frac{100+51}{100.51}\)

\(=\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+....+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\)

\(=\left(\frac{1}{100}+\frac{1}{99}+....+\frac{1}{52}+\frac{1}{51}\right)+\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)\)

\(=2\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)=2A\)

\(\Rightarrow \frac{A}{B}=\frac{151}{2}\)

30 tháng 4 2015

\(C=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{97.98}+\frac{1}{99.100}\)

\(C=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{99}-\frac{1}{100}\)

\(C=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{98}+\frac{1}{100}\right)\)

\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)

\(C=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(D=\frac{1}{51.100}+\frac{1}{52.99}+\frac{1}{53.98}+...+\frac{1}{99.52}+\frac{1}{100.51}\)

\(D=\frac{1}{151}.\left(\frac{151}{51.100}+\frac{151}{52.99}+\frac{151}{53.98}+...+\frac{151}{99.52}+\frac{151}{100.51}\right)\)

\(D=\frac{1}{151}.\left(\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+...+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\right)\)

\(D=\frac{1}{151}.\left(\frac{2}{100}+\frac{2}{99}+...+\frac{2}{51}\right)\)

\(D=\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)\)

\(\Rightarrow C:D=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)}\)

\(\Rightarrow C:D=\frac{151}{2}=75\frac{1}{2}\)

 

4 tháng 4 2016

Khó hiểu vậy ạ, giảng kĩ đc ko bạn :)

1 tháng 2 2018

Câu hỏi của Wang Jum Kai - Toán lớp 6 - Học toán với OnlineMath

7 tháng 4 2019

C:D ko phả STN nhé

6 tháng 8 2018

B= \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).....\left(1-\dfrac{1}{20}\right)\)

B= \(\dfrac{1}{2}.\dfrac{2}{3}.....\dfrac{19}{20}\)

B= \(\dfrac{1.2.....19}{2.3.....20}\)

B= \(\dfrac{1}{20}\)

19 tháng 4 2015

B=1/22+1/23+...+1/2100

2B=1/21+1/22+...1/299

2B-B=(1/21+1/22+...+1/299)-(1/22+1/23+...+1/2100)

B=1/21-1/2100=299/2100-1/2100=299-1/2100