Question

[Đỗ Văn Hải Đăng]

5/3*6+5/6*9+5/9*12+...+5/99*102 giúp mik với

Answer 1

[Ga'l wes]

\(\frac{5}{3\times6}+\frac{5}{6\times9}+\frac{5}{9\times12}+....+\frac{5}{99\times102}.\)

Đặt :

 \(A=\frac{5}{3\times6}+\frac{5}{6\times9}+\frac{5}{9\times12}+....+\frac{5}{99\times102}.\)

\(\frac{3}{5}\times A=\frac{3}{5}\times\left(\frac{5}{3\times6}+\frac{5}{6\times9}+...+\frac{5}{99\times102}\right)\)

\(\frac{3}{5}A=\frac{3}{3\times6}+\frac{3}{6\times9}+...+\frac{3}{96\times99}\)

\(\frac{3}{5}A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+....+\frac{1}{96}-\frac{1}{99}\)

\(\frac{3}{5}A=\frac{1}{3}-\frac{1}{99}\)

\(\frac{3}{5}A=\frac{32}{99}\)

\(A=\frac{32}{99}\div\frac{3}{5}=\frac{160}{297}\)

Answer 2

[Mai Anh Nguyen]

\(A=\frac{5}{3\times6}+\frac{5}{6\times9}+.....+\frac{5}{99\times102}\)

\(A=\frac{5}{3}\left[\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{9}\right)+.....+\left(\frac{1}{99}-\frac{1}{102}\right)\right]\)

\(A=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{102}\right)\)

\(A=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)

\(A=\frac{5}{3}.\frac{11}{34}\)

\(A=\frac{55}{102}\)