Question

Tìm x:

\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+.......+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)

Answer 1

Answer 2

Ta có : \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+....+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)

= \(\dfrac{1}{3}\) . ( \(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+....-\dfrac{1}{x+3}\)

=\(\dfrac{1}{3}\). ( \(\dfrac{1}{5}-\dfrac{1}{x+3}\)) = \(\dfrac{101}{1540}\)

=>\(\dfrac{1}{5}-\dfrac{1}{x+3}\) = \(\dfrac{303}{1540}\)

=> \(\dfrac{1}{x+3}\)= \(\dfrac{5}{1540}=\dfrac{1}{308}\)

=> x+3 = 308

=> x= 305

Vậy x= 305

Answer 3

\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)

\(\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x-3}\right)=\dfrac{101}{1540}\)

\(\dfrac{1}{5}-\dfrac{1}{x-3}=\dfrac{101}{1540}:\dfrac{1}{3}\)

Answer 4

còn lại bạn tự tính nhé