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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

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)

I don't now

mik ko biết 

sorry 

......................

25 tháng 7 2018

b,\(B=2^2+4^2+...+20^2\)

\(\Rightarrow B=2^2\left(1^2+2^2+...+10^2\right)\)

\(\Rightarrow B=4.\left[1.\left(2-1\right)+2.\left(3-1\right)+...+10.\left(11-1\right)\right]\)

\(\Rightarrow B=4\left(1.2-1+2.3-2+...+10.11-10\right)\)

\(\Rightarrow B=4\left[\left(1.2+2.3+...+10.11\right)-\left(1+2+...+10\right)\right]\)

\(\Rightarrow B=4\left(\frac{10.11.12}{3}-\frac{11.10}{2}\right)\)

24 tháng 5 2015

a)Ta có:

 A= 1/1.2+1/2.3+1/3.4+.....+1/99.100

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100

=99/100

b)Ta có:

B= 1/11+1/12+1/13+1/14+1/15+...+1/50

=(1/11+1/50)+(1/12+1/49)+...+(1/30+1/31)

=61/11.50+61/12.49+...+61/30.31

=61.(1/11.50+1/12.49+...+1/30.31)

Mình xin lỗi chỉ làm được đến đây vì dạng tính B mình không tốt lắm ◕◡◕

 

24 tháng 5 2015

\(B=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{50}\right)>\left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)=> \(B>\frac{20}{30}+\frac{20}{50}=\frac{2}{3}+\frac{2}{5}=\frac{16}{15}>1\)

mà \(A=\frac{99}{100}