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

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

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

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

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

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

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

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

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

S = 2(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

\(A=\frac{10^2}{1\cdot6}+\frac{10^2}{6\cdot11}+...+\frac{10^2}{61\cdot66}=\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{61\cdot66}\right)\cdot20\)

\(=\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{61}-\frac{1}{66}\right)\cdot20\)

\(=\left[\left(1-\frac{1}{66}\right)+\left(\frac{1}{6}-\frac{1}{6}\right)+...+\left(\frac{1}{61}-\frac{1}{61}\right)\right]\cdot20\)

\(=\left[\left(\frac{66}{66}-\frac{1}{66}\right)+0+...+0\right]\cdot20=\frac{65}{66}\cdot20=\frac{65\cdot20}{66}=\frac{65\cdot10}{33}=\frac{650}{33}\)

\(A=\frac{10^2}{1.6}+\frac{10^2}{6.11}+...+\frac{10^2}{61.66}\)

\(=10^2.\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{61.66}\right)\)

\(=10^2.5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{61}-\frac{1}{66}\right)\)

\(=500.\left(1-\frac{1}{66}\right)\)

\(=500.\frac{65}{66}\)

\(=\frac{16250}{33}\)

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

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

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

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

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

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

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

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

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

\(S=\frac{10}{1.6}+\frac{10}{6.11}+...+\frac{10}{101.106}\)

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

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

\(=2\left(1-\frac{1}{106}\right)=2.\frac{105}{106}=\frac{105}{53}\)

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

Đặt \(A=\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{46\cdot51}\)

\(5A=5\left(\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{46\cdot51}\right)\)

\(5A=\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{5}{46\cdot51}\)

\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{46}-\frac{1}{51}\)

\(5A=1-\frac{1}{51}\)

\(5A=\frac{50}{51}\Rightarrow A=\frac{50}{51}:5\Rightarrow A=\frac{10}{51}\)

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

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

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

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

Cảm ơn nhé

Thank you