\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{99}{100}\)
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21 tháng 4 2021
1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)
\(=0\)
12 tháng 5 2022
a: \(=\dfrac{99}{100}:\left(\dfrac{3}{12}-\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{49}{25}\)
\(=\dfrac{99}{100}:\dfrac{1}{2}-\dfrac{49}{25}\)
\(=\dfrac{99}{50}-\dfrac{98}{50}=\dfrac{1}{50}\)
b: \(=\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{39}{60}+\dfrac{-19}{60}\cdot\dfrac{24}{47}\)
=459/940
R
30 tháng 3 2018
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
13 tháng 5 2022
9: \(=1-\dfrac{1}{99}+1-\dfrac{1}{100}+\dfrac{100}{101}\cdot\dfrac{1-4+3}{12}=2-\dfrac{199}{9900}=\dfrac{19601}{9900}\)
10: \(=\left(\dfrac{78}{79}+\dfrac{79}{80}+\dfrac{80}{81}\right)\cdot\dfrac{6+5+9-20}{30}=0\)
31 tháng 5 2021
Ta có \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}=\dfrac{1}{6}-\dfrac{1}{6}=0\) nên Q = 0.
DT
Đỗ Thanh Hải
CTVVIP
29 tháng 5 2017
a) Ta có
S = \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)2S = \(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)
S = \(\dfrac{1}{4}-\dfrac{1}{\left(n+1\right).\left(n+2\right):2}\)
b) A = \(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}\)
A = \(2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
A = \(2-\dfrac{1}{99}\)
A = \(\dfrac{197}{99}\)
c) Ta có
B = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
B = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B = \(1-\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
d) Ta có
C = \(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
C = \(1+\left(1+\dfrac{98}{2}\right)+\left(1+\dfrac{97}{3}\right)+...+\left(1+\dfrac{1}{99}\right)\)
C = \(1+50+\dfrac{100}{3}+...+\dfrac{100}{99}\)
C = 51 + 100(\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\))
Đặt D = \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{99}\)
D = \(\dfrac{97}{198}\)
=> C = 51 + 100.\(\dfrac{97}{198}\)
C = 51 + \(\dfrac{4850}{99}\)
C = \(\dfrac{9899}{99}\)
Đây là bài làm của mình sai thì nx nha
19 tháng 6 2017
\(\dfrac{100+\dfrac{99}{2}+\dfrac{98}{3}+...+\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\left(\dfrac{99}{2}+1\right)+\left(\dfrac{98}{3}+1\right)+...+\left(\dfrac{1}{100}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\dfrac{101}{2}+\dfrac{101}{3}+...+\dfrac{101}{100}+\dfrac{101}{101}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{101\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
\(=101-2\)
\(=99\)
\(\Leftrightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{99}{100}\)
\(\Leftrightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{99}{100}\)
\(\Leftrightarrow1-\dfrac{1}{n+1}=\dfrac{99}{100}\)
\(\Leftrightarrow\dfrac{1}{n+1}=\dfrac{1}{100}\)
=> n+1=100
n=99
1/2 + 1/6 + 1/12 + ... + 1/(n(n+1)) = 99/100`
`=> 1/(1.2) + 1/(2.3) + 1/(3.4) + ... + 1/(n.(n+1)) = 99/100`
`=> 1/1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 + ... + 1/n - 1/(n+1) = 99/100`
`=> 1 - 1/(n+1) = 99/100`
`=> 1/(n + 1) = 1 - 99/100`
`=> 1/(n + 1 ) = 1/100`
`=> n +1 = 100`
`=> n = 99`