Question

[Dragonball songoku]

Tính

\(\frac{6}{18}\)+\(\frac{6}{54}\)+\(\frac{6}{108}\)+...+\(\frac{6}{990}\)

Giúp mình với nhé!

Answer 1

[Hoang Thien Duc]

Ta làm như sau:

   \(\frac{6}{18}\)+\(\frac{6}{54}\)+\(\frac{6}{108}\)+...+\(\frac{6}{990}\)

=\(\frac{6}{3.6}\)+\(\frac{6}{6.9}\)+\(\frac{6}{9.12}\)+...\(\frac{6}{30.33}\)

=2 (\(\frac{3}{3.6}\)+\(\frac{3}{6.9}\)+\(\frac{3}{9.12}\)+...+\(\frac{3}{30.33}\)

=2 (\(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\))

=2 ( \(\frac{1}{3}-\frac{1}{33}\))

=2.\(\frac{10}{33}\)=\(\frac{2.10}{33}\)=\(\frac{20}{33}\)

Answer 2

[Đinh Gia Bảo]

\(\frac{6}{18}+\frac{6}{54}+\frac{6}{108}+...+\frac{6}{990}\)

=\(\frac{6}{3.6}+\frac{6}{6.9}+\frac{6}{9.12}+...+\frac{6}{30.33}\)

= 2.(\(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\))

=2.(\(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\))

=2.[\(\frac{1}{3}+\left(\frac{-1}{6}+\frac{1}{6}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)+...+\left(\frac{-1}{30}+\frac{1}{30}\right)+\frac{-1}{33}\)]

=2.\(\left[\frac{1}{3}+\frac{-1}{33}\right]\)

=2.\(\left[\frac{11}{33}+\frac{-1}{33}\right]\)

=2.\(\frac{10}{33}\)

=\(\frac{20}{33}\)