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\(A=2\left(\dfrac{1}{6}+\dfrac{1}{66}+...+\dfrac{1}{\left(5n-4\right)\left(5n+1\right)}\right)\)

\(=2\left(\dfrac{1}{1\cdot6}+\dfrac{1}{6\cdot11}+...+\dfrac{1}{\left(5n-4\right)\left(5n+1\right)}\right)\)

\(=\dfrac{2}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n-4\right)\left(5n+1\right)}\right)\)

\(=\dfrac{2}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{5n-4}-\dfrac{1}{5n+1}\right)\)

\(=\dfrac{2}{5}\left(1-\dfrac{1}{5n+1}\right)=\dfrac{2}{5}\cdot\dfrac{5n+1-1}{5n+1}\)

\(=\dfrac{2}{5}\cdot\dfrac{5n}{5n+1}=\dfrac{2n}{5n+1}\)

28 tháng 3

15 tháng 4 2019

Ta có:\(\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)

        \(=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)

        \(=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)

        \(=\frac{1}{5}.\left(1-\frac{1}{5n+6}\right)\)

        \(=\frac{1}{5}.\left(\frac{5n+5}{5n+6}\right)=\frac{n+1}{5n+6}\left(\text{đ}pcm\right)\)

18 tháng 2 2017

a) 51^2k = (51^2)^k = ....01^k= ...01

9 tháng 8 2019

Ta có : \(A=\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{(5n+1)(5n+6)}\)

\(=\frac{1}{5}\cdot\left[\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{5}{(5n+1)(5n+6)}\right]\)

\(=\frac{1}{5}\cdot\left[1-\frac{1}{5n+6}\right]=\frac{1}{5}\cdot\frac{5n+6-1}{5n+6}=\frac{1}{5}\cdot\frac{5(n+1)}{5n+6}=\frac{n+1}{5n+6}\)

26 tháng 7 2020

Cho:
m-n+p-q \vdots 3
2m+2n+2p-2q \vdots 4
-m-3n+p-3q \vdots -6
6m+8n+2p-6q \vdots 5
Hãy tính:
\frac{(2m-3q)^6+(5n-p)^4}{(9m+5n-4p+6q)^2}=?
A.\frac{1}{75000}
B.\frac{1}{75076}
C.\frac{1}{80000}
D.\frac{1}{85076}

5 tháng 4 2015

\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{\left(5n+1\right).\left(5n+6\right)}\)

\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\)

\(5A=1-\frac{1}{5n+6}=\frac{5n+6-1}{5n+6}=\frac{5n+5}{5n+6}\)=> \(A=\frac{n+1}{5n+6}\)