A = 1 / 6x7 + 1/7.8 + 1/8.9 + . . . + 1/2017x2018
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28 tháng 6 2015
1/9.10 - ( 1/1.2 + 1/2.3 +...+ 1/8 .9)
= 1/9 - 1/10 -( 1/1 - 1/2 + 1/2 - 1/3 +. ..+1/8 - 1 9/)
= 1/9 - 1/10 - ( 1 - 1/9 )
= 1/9 - 1/10 - 1 + 1/9
=-79/9
29 tháng 6 2015
thang Tran làm cách giải đúng rồi, nhưng đáp số là -79/90 nha.
Dù sao cũng cám ơn bạn vì giúp mình hiểu bài.
BU
3
24 tháng 2 2016
A= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/ 7.8 + 1/ 8.9 + 1/ 9.10
=> A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10
=> A = 1 - 1/10 = 9/10
Vậy A = 9/10
24 tháng 2 2016
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10
A = 1 - 1/10 = 9/10
9 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}-\dfrac{1}{5\cdot6}-\dfrac{1}{6\cdot7}-\dfrac{1}{7\cdot8}-\dfrac{1}{8\cdot9}\)
`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
`=`\(\dfrac{1}{3}-\left(\dfrac{1}{2}-\dfrac{1}{9}\right)\)
`=`\(\dfrac{1}{3}-\dfrac{7}{18}=-\dfrac{1}{18}\)
21 tháng 5 2022
\(=\dfrac{1}{9}-\dfrac{1}{10}-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=\dfrac{1}{9}-\dfrac{1}{10}-\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{1}{9}-\dfrac{1}{10}-1+\dfrac{1}{9}\)
\(=-\dfrac{97}{90}\)
25 tháng 2 2018
\(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}\)
\(=\frac{8-7}{7\cdot8}+\frac{9-8}{8\cdot9}+\frac{10-9}{9\cdot10}+\frac{11-10}{10\cdot11}+\frac{12-11}{11\cdot12}+\frac{13-12}{12\cdot13}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{7}-\frac{1}{13}=\frac{13-7}{7\cdot13}=\frac{6}{91}\)
LV
0
16 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)
`=`\(\dfrac{2}{9}\)
Vậy, \(A=\dfrac{2}{9}\)
`b)`
\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)
Vậy, \(B=\dfrac{4}{25}\)
`c)`
\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)
`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
Vậy, \(C=\dfrac{99}{100}\)
\(A=\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{2017\cdot2018}\)
\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{2017}-\frac{1}{2018}\)
\(A=\frac{1}{6}-\frac{1}{2018}\)
\(A=\frac{503}{3027}\)
Vậy ...............................................
\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{2017.2018}\)
\(\implies A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(\implies A=\frac{1}{6}-\frac{1}{2018}\)
\(\implies A=\frac{503}{3027}\).
~ Hok tốt a~