\(\frac{5}{11x12}+\frac{5}{12x13}+...+\frac{5}{98x99}\)
=\(\frac{5}{11}-\frac{5}{12}+\frac{5}{12}-\frac{5}{13}+...+\frac{5}{98}-\frac{5}{99}\)
=\(\frac{5}{11}-\frac{5}{99}\)
=\(\frac{40}{99}\)
Cái cuối bỏ 1 số 0 thì đúng hơn nha bạn
\(\frac{5}{11.12}+\frac{5}{12.13}+\frac{5}{13.14}+...+\frac{5}{98.99}\)
\(=5\left(\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+...+\frac{1}{98.99}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{99}\right)\)
\(=5.\frac{8}{99}\)
\(=\frac{40}{99}\)
\(=5\left(\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+...+\frac{1}{98.99}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{99}\right)=5.\frac{8}{99}=\frac{40}{99}\)
\(\frac{5}{11.12}+\frac{5}{12.13}+\frac{5}{13.14}+.........+\frac{5}{98.99}\)
\(=5\left(\frac{1}{11.12}+\frac{1}{12.13}+........+\frac{1}{98.99}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+....+\frac{1}{98}-\frac{1}{99}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{99}\right)\)
\(=5.\frac{8}{99}=\frac{40}{99}\)