5S=5.(1/1.6+1/6.11+...+1/496.501)

5S=5/1.6+5/6.11+...+5/496.501

5S=1/1-1/6+1/6-1/11+...+1/496-1/501

5S=1-1/501

5S=500/501

S=500/501:5=100/501

k nhé

ta co:5S=5/1.6+5/6.11+5/11.16+...+5/496.501

             =1-1/6+1/6-1/11+1/11-1/16+.....+1/496-1/501

             =1-1/501=500/501

       =>S=500/501:5=100/501

MK đau tien nha bn

1/1.6 + 1/6.11+ 1/11.16+ ....  

số thứ 100 có dạng 1/(496.501)  

do đó tổng trên bằng :

1/5( 1/1- 1/501)

= 100/ 501

1/1-1/6+1/6-1/11+...+1/496-1/501

=1/1-1/501=500/501

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

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

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

=1/5(5/1*6+5/6*11+...+5/101*106)

=1/5(1-1/6+1/6-1/11+...+1/101-1/106)

=1/5(1-1/106)

=1/5*105/106

=21/106

\(B=\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{101.106}\)

\(B=\dfrac{1}{5}.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\)

\(B=\dfrac{1}{5}.\left(1-\dfrac{1}{106}\right)\)

\(B=\dfrac{1}{5}.\dfrac{105}{106}\)

\(B=\dfrac{21}{106}\)

Giải:

a) S=5²/1.6+5²/6.11+5²/11.16+5²/16.21+5²/21.26

    S=5.(5.1/6+5/6.11+5/11.16+5/16.21+5/21.26)

    S=5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21+1/21-1/26)

    S=5.(1/1-1/26)

    S=5.25/26

    S=125/26

b) (1-1/2).(1-1/3).(1-1/4).(1-1/5).....(1-1/19).(1-1/20)

=1/2.2/3.3/4.4/5.....18/19.19/20

=1.2.3.4.....18.19/2.3.4.5.....19.20

=1/20

Chúc bạn học tốt!

Lời giải:
\(5A=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{501-496}{496.501}\)

\(=\frac{6}{1.6}-\frac{1}{1.6}+\frac{11}{6.11}-\frac{6}{6.11}+\frac{16}{11.16}-\frac{11}{11.16}+...+\frac{501}{496.501}-\frac{496}{496.501}\)

\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+....+\frac{1}{496}-\frac{1}{501}=1-\frac{1}{501}=\frac{500}{501}\)

$\Rightarrow A=\frac{100}{501}$

\(A=\dfrac{1}{5}\left(\dfrac{1}{1.6}+...+\dfrac{1}{496.501}\right)\)

\(A=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\cdot\cdot\cdot+\dfrac{1}{495}-\dfrac{1}{501}\right)\)

\(A=\dfrac{1}{5}\left(1-\dfrac{1}{501}\right)\)

\(A=\dfrac{1}{5}\cdot\dfrac{500}{501}=\dfrac{100}{501}\)

\(A=\dfrac{1}{1\cdot6}-\dfrac{1}{6\cdot11}-\dfrac{1}{11\cdot16}-\dfrac{1}{16\cdot21}-...-\dfrac{1}{46\cdot51}\)

\(=\dfrac{1}{6}-\left(\dfrac{1}{6\cdot11}+\dfrac{1}{11\cdot16}+\dfrac{1}{16\cdot21}+...+\dfrac{1}{46\cdot51}\right)\)

\(=\dfrac{1}{6}-\dfrac{1}{5}\left(\dfrac{5}{6\cdot11}+\dfrac{5}{11\cdot16}+\dfrac{5}{16\cdot21}+...+\dfrac{5}{46\cdot51}\right)\)

\(=\dfrac{1}{6}-\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{46}-\dfrac{1}{51}\right)\)

\(=\dfrac{1}{6}-\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{51}\right)\)

\(=\dfrac{1}{6}-\dfrac{1}{5}\cdot\dfrac{5}{34}\)

\(=\dfrac{1}{6}-\dfrac{1}{34}\)

\(=\dfrac{7}{51}\)

Vậy \(A=\dfrac{7}{51}\)