tk mị̣̣̉̉̉̉̉̉̀̉̃́́́nh nhe !

câu hỏi tương tự có đó bạn, bạn vào tham khảo nhe!

Ta có

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

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

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

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

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

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

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

thanks bn nhìu mik cũng nghĩ vậy đó

lỗi

mik sửa r nhé

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

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

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

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

\(=\dfrac{n+1}{5n+6}=VP\)

\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{10000}\right)\)

\(=\left(\frac{4}{4}-\frac{1}{4}\right).\left(\frac{9}{9}-\frac{1}{9}\right)...\left(\frac{10000}{10000}-\frac{1}{10000}\right)\)

\(=\frac{3}{4}.\frac{8}{9}...\frac{9999}{10000}=\frac{3}{2.2}.\frac{2.4}{3.3}...\frac{99.101}{100.100}\)

\(=\frac{101}{100}\)

\(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\)

\(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)

\(=5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(=5.\left(\frac{1}{1}-\frac{1}{31}\right)=5.\left(\frac{31}{31}-\frac{1}{31}\right)=5.\frac{30}{31}=\frac{150}{31}\)

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 

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

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

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

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

=> A  = \(=\frac{5\left(n+1\right)}{5n+6}:5=\frac{5\left(n+1\right)}{5n+6}\cdot\frac{1}{5}=\frac{n+1}{5n+6}\)

VẬy VT = VP ĐT Đ CM