\(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + .... + \(\dfrac{1}{99.101}\)
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Những câu hỏi liên quan
13 tháng 2 2022
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\left(\dfrac{100}{101}\right)=\dfrac{50}{101}\)
13 tháng 2 2022
\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)
6 tháng 5 2018
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
6 tháng 5 2018
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
3 tháng 3 2017
2/ = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +......+\(\dfrac{1}{100.101}\)
= 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)+.........+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)
=1-\(\dfrac{1}{101}\)=...........
mk làm vậy thôi nha
thông cảm
LB
2 tháng 3 2017
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}\)=\(1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}\)
=1-\(\dfrac{1}{5}=\dfrac{4}{5}\)
tương tự
NH
11 tháng 3 2023
\(B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\\ B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\\ B=\dfrac{1}{1}-\dfrac{1}{101}\\ B=\dfrac{101}{101}-\dfrac{1}{101}\\ B=\dfrac{100}{101}\)
NH
11 tháng 3 2023
\(\dfrac{2}{1\cdot3}=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3}{3}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\dfrac{2}{3\cdot5}=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)
\(\dfrac{2}{5\cdot7}=\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{7}{35}-\dfrac{5}{35}=\dfrac{2}{35}\)
và cứ như thế đến số cuối
VN
2
VT
3 tháng 3 2023
\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\\ =\dfrac{4}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =2.\left(1-\dfrac{1}{101}\right)\\ =2.\dfrac{100}{101}\\ =\dfrac{200}{101}\)
25 tháng 1 2023
\(B=-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{-1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{100}{101}=-\dfrac{50}{101}\)
30 tháng 6 2017
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...\dfrac{1}{99}-\dfrac{1}{101}\right)-\dfrac{1}{101}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{101}\right)-\dfrac{1}{101}\)
\(A=\dfrac{1}{2}.\left(\dfrac{100}{101}\right)-\dfrac{1}{101}\)
\(A=\dfrac{50}{101}-\dfrac{1}{101}=\dfrac{49}{101}\)
30 tháng 6 2017
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{99.101}-\dfrac{1}{101}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)-\dfrac{1}{101}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)-\dfrac{1}{101}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)-\dfrac{1}{101}\)
\(=\dfrac{1}{2}.\dfrac{100}{101}-\dfrac{1}{101}=\dfrac{50}{101}-\dfrac{1}{101}=\dfrac{49}{101}\)
Vậy...
TC
31 tháng 3 2017
a)\(\dfrac{1}{2}\).(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+........+\(\dfrac{2}{99.101}\))
=\(\dfrac{1}{2}\).(1-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+........+\(\dfrac{1}{99}\)-\(\dfrac{1}{101}\))
=\(\dfrac{1}{2}\).(1-\(\dfrac{1}{101}\))
=\(\dfrac{1}{2}\).\(\dfrac{100}{101}\)
=\(\dfrac{100}{202}\)=\(\dfrac{50}{101}\)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}.\dfrac{100}{101}=\dfrac{50}{101}\)