Bài 5: Tính nhanh
A = 1/1 x3 + 1/ 3 x 5 + 1/ 5 x 7 + ................. + 1/ 2013 x 2015
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19 tháng 4 2021
Câu 1:
a) \(-\dfrac{3}{7}-\left(\dfrac{2}{3}-\dfrac{3}{7}\right)=\dfrac{-3}{7}-\dfrac{2}{3}+\dfrac{3}{7}=\dfrac{-2}{3}\)
19 tháng 4 2021
Câu 2:
b) \(\dfrac{2}{15}:\left(\dfrac{1}{3}\cdot\dfrac{4}{5}-\dfrac{1}{3}\cdot\dfrac{6}{5}\right)=\dfrac{2}{15}:\left[\dfrac{1}{3}\left(\dfrac{4}{5}-\dfrac{6}{5}\right)\right]=\dfrac{2}{15}:\left(\dfrac{1}{3}\cdot\dfrac{-2}{5}\right)=\dfrac{2}{15}:\dfrac{-2}{15}=\dfrac{2}{-2}=-1\)
9 tháng 7 2018
\(a)\) \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3S=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3S-S=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(2S=3+\frac{1}{3^7}\)
\(2S=\frac{3^8+1}{3^7}\)
\(S=\frac{3^8+1}{3^7}.\frac{1}{2}\)
\(S=\frac{3^8+1}{2.3^7}\)
Vậy \(S=\frac{3^8+1}{2.3^7}\)
Chúc bạn học tốt ~
17 tháng 6 2021
Giải:
a) \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)
\(=\dfrac{1.2.3.4}{2.3.4.5}\)
\(=\dfrac{1}{5}\)
b) \(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\)
\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\)
\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\)
\(=\dfrac{1}{100}\)
Chúc bạn học tốt!
2
21 tháng 12 2022
`a)(1-1/2)xx(1-1/3)xx(1-1/4)xx(1-1/5)`
`=1/2xx2/3xx3/4xx4/5`
`=[1xx2xx3xx4]/[2xx3xx4xx5]`
`=1/5`
`b)(1-3/4)xx(1-3/7)xx(1-3/10)xx(1-3/13)xx .... xx(1-3/97)xx(1-3/100)`
`=1/4xx4/7xx7/10xx10/13xx .... xx94/97xx97/100`
`=[1xx4xx7xx10xx...xx94xx97]/[4xx7xx10xx13xx....xx97xx100]`
`=1/100`
TV
2
31 tháng 8 2016
\(\frac{x-1}{2015}+\frac{x-3}{2013}=\frac{x-5}{2011}+\frac{x-7}{2009}\)
=> \(\frac{x-1}{2015}-1+\frac{x-3}{2013}-1=\frac{x-5}{2011}-1+\frac{x-7}{2009}-1\)
=> \(\frac{x-2016}{2015}+\frac{x-2016}{2013}=\frac{x-2016}{2011}+\frac{x-2016}{2009}\)
=> \(\frac{x-2016}{2009}+\frac{x-2016}{2011}-\frac{x-2016}{2013}-\frac{x-2016}{2015}=0\)
=> \(\left(x-2016\right).\left(\frac{1}{2009}+\frac{1}{2011}-\frac{1}{2013}-\frac{1}{2015}\right)\)
Vì \(\frac{1}{2009}>\frac{1}{2013};\frac{1}{2011}>\frac{1}{2015}\)
=> \(\frac{1}{2009}+\frac{1}{2011}-\frac{1}{2013}-\frac{1}{2015}\ne0\)
=> \(x-2016=0\)
=> \(x=2016\)
A x 2 = 2/1 x3 + 2/ 3 x 5 + 2/ 5 x 7 + ................. + 2/ 2013 x 2015
= 1/1 – 1/3 + 1/3 – 1/5 + 1/5 – 1/7 + .................. + 1/2013 – 1/2015
= 1 – 1/2015 = 2014/2015
Vậy A = 2014/2015 : 2 = 2014/4030.
\(\frac{2014}{4030}\)