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Ta có: \(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1}{a\left(a+1\right)}-\dfrac{a}{a\left(a+1\right)}\)
\(=\dfrac{a+1-a}{a\left(a+1\right)}\)
\(=\dfrac{1}{a\left(a+1\right)}\) (đpcm)
Ta được:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+...-\dfrac{1}{100}\) \(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
a: \(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}=1-\dfrac{1}{2008}=\dfrac{2007}{2008}\)
b: \(Q=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2009\cdot2011}\right)\)
\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{7}{2}\cdot\dfrac{2010}{2011}\simeq3,50\)
a, A = 1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/2017 - 1/2018
A = 1 - 1/2018 = 2017/2018
b, B = 5/2 . ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2016 -1/2018)
B= 5/2 . ( 1/2 - 1/ 2018 )
B = 504/1009
c, C = 1/3.6 + 1/ 6.9 + 1/ 9.12 + ... + 1/ 30.33
C= 1/3 - 1/6 + 1/6 - 1/ 9 + 1/9 - 1/12 + ... + 1/30 - 1/33
C = 1/3 - 1/33
C= 10/33
phan B mk quên nhân với 5/2
lấy 5/2 . 504/1009 = 1260/1009
A = \(\dfrac{9}{1.2}\)+ \(\dfrac{9}{2.3}\)+\(\dfrac{9}{3.4}\)+......+\(\dfrac{99}{99.100}\)
A = 9( \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.......+\(\dfrac{1}{99.100}\))
A = 9( 1-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+........+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\))
A = 9 ( 1 - \(\dfrac{1}{100}\))
A = 9 . \(\dfrac{99}{100}\)
A = \(\dfrac{891}{100}\)
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{98\cdot99}+\dfrac{9}{99\cdot100}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=9\left(1-\dfrac{1}{100}\right)\)
\(=9\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)
\(=9\cdot\dfrac{99}{100}\)
\(=\dfrac{891}{100}\)
a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\dfrac{1}{1}+0+0+...+0-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
a) 11.2+12.3+13.4+....+199.10011.2+12.3+13.4+....+199.100
= 11−12+12−13+13−14+....+199−110011−12+12−13+13−14+....+199−1100
=11+0+0+...+0−110011+0+0+...+0−1100
=1−11001−1100
= 99100
1,
B=\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.........+\(\dfrac{1}{2^{2017}}\)
2B=1+\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+.......+\(\dfrac{1}{2^{2016}}\)
2B-B=(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+.......+\(\dfrac{1}{2^{2016}}\))-(\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.......+\(\dfrac{1}{2^{2017}}\))
B=1-\(\dfrac{1}{2^{2017}}\)
Vậy B=1-\(\dfrac{1}{2^{2017}}\)
\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)
= \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{8}\) MSC: 8
= \(\dfrac{4}{8}\) + \(\dfrac{1}{8}\)
= \(\dfrac{5}{8}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
= \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
= \(\dfrac{1}{2}-\dfrac{1}{8}\)
=\(\dfrac{4}{8}-\dfrac{1}{8}\)
=\(\dfrac{3}{8}\)
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...............+\dfrac{2}{2008.2009}\)
\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+................+\dfrac{1}{2008.2009}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.................+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)
\(=2\left(1-\dfrac{1}{2009}\right)\)
\(=2.\dfrac{2008}{2009}=\dfrac{4016}{2009}\)
Sửa lại đề:
\(M=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{49.50}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-..........-\dfrac{1}{49}-\dfrac{1}{50}\)
\(M=1-\dfrac{1}{50}\)
\(M=\dfrac{50}{50}-\dfrac{1}{50}\)
\(M=\dfrac{49}{50}\)
Đề bài là thu gọn / tính giá trị biểu thức nhé chứ không phải là So sánh , thiếu dữ kiện kìa
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{2022.2023}\\ \\ \\ \\ \\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{1}{1}-\dfrac{1}{2023}=\dfrac{2023}{2023}-\dfrac{1}{2023}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2022}{2023}\)