\(A=\frac{20}{1\cdot6}+\frac{20}{6\cdot11}+...+\frac{20}{51\cdot56}+\frac{20}{56\cdot61}\)

\(A=\frac{20}{5}\cdot\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{51}-\frac{1}{56}+\frac{1}{56}-\frac{1}{61}\right)\)

\(A=4\cdot\left(1-\frac{1}{61}\right)\)

\(A=4\cdot\frac{60}{61}\)

\(A=\frac{240}{61}\)

\(A=\frac{20}{1.6}+\frac{20}{6.11}+...+\frac{20}{51.56}+\frac{20}{56.61}\)

\(A=\frac{20}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{51}-\frac{1}{56}+\frac{1}{56}-\frac{1}{61}\right)\)

\(A=4.\left(1-\frac{1}{61}\right)\)

\(A=4.\frac{60}{61}=\frac{240}{61}\)

\(S=2\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{101.106}\right)\)

\(S=2\left(1-\frac{1}{106}\right)\)

\(S=\frac{210}{106}=\frac{105}{53}\)

\(S=2.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{101.106}\right)\)

\(S=2.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-...-\frac{1}{101}+\frac{1}{101}-\frac{1}{106}\right)\)

\(S=2.\left[1+\left(\frac{-1}{6}+\frac{1}{6}\right)+\left(\frac{-1}{11}\frac{1}{11}\right)+...+\left(\frac{-1}{101}+\frac{1}{101}\right)-\frac{1}{106}\right]\)

\(S=2.\left[1+0+0+...+0-\frac{1}{106}\right]\)

\(S=2.\left[1-\frac{1}{106}\right]\)

\(S=2.\frac{105}{106}\)

S = 2(5/1.6 + 5/6.11 +.......+ 5/101.106)

S = 2( 1 - 1/6 + 1/6 - 1/11 +.....+ 1/101 - 1/106)

S = 2( 1 - 1/106)

S = 2 . 105/106

S = 105/53

k mk đi,mk mới bị trừ điểm!

1/2.S=5/(1.6)+5/(6.11)+...+5/(101.106)

1/2.S=1/1-1/6+1/6-1/11+...+1/101-1/106

1/2.S=1/1-1/106

1/2.S=105/106

S=105/53

Ta có :

\(\frac{5}{1.6}+\frac{5}{6.11}+................+\frac{5}{\left(5.x+1\right).\left(5.x+6\right)}=\)\(\frac{50}{41}\)

=> \(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...............+\frac{1}{5.x+1}-\frac{1}{5.x+6}\) = \(\frac{50}{41}\)

=> \(1-\frac{1}{5.x+6}=\frac{50}{41}\)

=> \(\frac{1}{5.x+6}=\frac{-9}{41}\)................ mình ko tìm ra vì p/s kia ko có tử là 1

bạn xem lại đề bài giúp mình nha 

\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{46.51}\)

\(=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{46.51}\right)\)

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

\(=\frac{1}{5}.\left(1-\frac{1}{51}\right)\)

\(=\frac{1}{5}.\left(\frac{51}{51}-\frac{1}{51}\right)\)

\(=\frac{1}{5}.\frac{50}{51}\)

\(=\frac{10}{51}\)

Chúc bạn học tốt !!! 

=1/5.(1-1/51)

=1/5.50/51

=10/51

\(\Leftrightarrow B=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(\Leftrightarrow B=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)\)

\(\Leftrightarrow B=\frac{3}{5}.\frac{100}{101}\)

\(\Leftrightarrow B=\frac{60}{101}\)

Cảm ơn nhé

Thank you

1. a)                                                      giai                                                                                                                       ta co (1/2.5=1/2-1/5)+(1/5.8=1/5-1/8)+....+(1/2009.2012=1/2009-1/2012)                                                                 =>a=1/2-1/5+1/5-1/8+...+1/2009-1/2012                                                                                                                    <=>1/2-1/2012                                                                                                                                                          =>a=1005/2012                câu b bằng nhau nhhung minh không th
2. e giải ra được

a,1/1-1/4+1/4-1/7+...+1/2008-1/2011

=(1-1/2011)+(-1/4+1/4)+...+(-1/2008+1/2008)

=1-1/2011+0+...+0

=1-1/2011

=2010/2011