\(\frac{2}{1.3}\) + \(\f...">
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23 tháng 1 2017

a, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

=2.(\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\))

=\(2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

=\(\frac{2}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{100}{101}\)

b, \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)

=\(5.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)

=\(5.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)

=\(\frac{250}{101}\)

\(=\frac{5}{2}.\frac{100}{101}\)

3 tháng 5 2019

a,21.321.3+23.523.5+25.725.7+....+299.101

=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)

=>\(\frac{1}{1}-\frac{1}{101}\)

=>\(\frac{100}{101}\)

b,

51.351.3+53.553.5+55.755.7+....+599.101

=>\(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{99.101}\right)\)

=>\(\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)

=>\(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)

=>\(\frac{5}{2}.\frac{100}{101}\)

=>\(\frac{250}{101}\)

2 tháng 5 2018

\(A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}+\frac{2}{99.101}\)

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)

\(A=1-\frac{1}{101}\)

\(A=\frac{101}{101}-\frac{1}{101}\)

\(A=\frac{100}{101}\)

Chúc bạn học tốt !!! 

2 tháng 5 2018

A = 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/99 - 1/101 

A = 1/1 - 1/101 

A = 101/101 - 1/101 

A = 100/101 

6 tháng 5 2018

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+...\)\(+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

6 tháng 5 2018

TA ĐẶT: \(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)

\(2A=\frac{2\cdot1}{1\cdot3\cdot2}+\frac{2\cdot1}{3\cdot5\cdot2}+...+\frac{2\cdot1}{99\cdot101\cdot2}\)

\(2A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)

\(2A=1\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{5}+...+\frac{1}{99}\cdot\frac{1}{101}\)

\(2A=1\cdot\frac{1}{101}=\frac{1}{101}\)

\(A=\frac{1}{101}:2=\frac{1}{202}\)

CHẮC LÀ ĐÚNG ĐÓ BN. CHÚC BN HOK TỐT. ^_^

22 tháng 2 2017

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

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

\(A=\frac{99}{100}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(B=2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(B=1.\left(1-\frac{1}{101}\right)\)

\(B=\frac{100}{101}\)

\(C=\frac{4}{4.7}+\frac{4}{7.10}+\frac{4}{10.13}+...+\frac{4}{73.76}\)

\(C=4.\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)

\(C=4.\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)

\(C=\frac{4}{3}.\left(\frac{1}{4}-\frac{1}{76}\right)\)

\(C=\frac{4}{3}.\frac{9}{38}\)

\(C=\frac{6}{19}\)

22 tháng 2 2017

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\\ =\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+....+\left(\frac{1}{99}-\frac{1}{100}\right)\\ =1-\frac{1}{100}\\ =\frac{99}{100}\\ B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\\ =\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\\ =\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

\(C=\frac{4}{4.7}+\frac{4}{7.10}+....+\frac{4}{73.76}\\ =\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{73.76}\right)\\ =\frac{4}{3}.\left(\frac{3}{4}-\frac{3}{76}\right)\\ =\frac{18}{19}\)

Học tốt Nghe!!

2 tháng 4 2018

2/7A=2/1.3+2/3.5+...+2/99+101

2/7A=1-1/3+1/3-1/5+...+1/99-1/101

2/7A=1-1/101

2/7A=100/101

A=350/101

B=(1+1+1+1)-(1/2+1/6+1/12+1/20)

=4-(1/1.2+1/2.3+1/3.4+1/4.5)

=4-(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5)

=4-(1-1/5)

=4-4/5

=16/5

8 tháng 5 2017

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{101}\right)\)

\(A=\frac{1}{2}.\frac{100}{101}\)

\(A=\frac{50}{101}\)

\(A=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)

\(A=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+...+\frac{3^2}{17.20}\)

\(A=\frac{3^2}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)

\(A=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(A=3\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(A=3.\frac{9}{20}\)

\(A=\frac{27}{20}\)

k nhá bn!

8 tháng 5 2017

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{5}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(2A=1-\frac{1}{101}\)

\(2A=\frac{100}{101}\)

\(\Rightarrow A=\frac{50}{101}\)

\(A=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)

\(A=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+...+\frac{3^2}{17.20}\)

\(\Rightarrow A=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\right)\)

\(A=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(A=3\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(A=3.\frac{9}{20}\)

\(A=\frac{27}{20}\)