Question

P = \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+.....+\dfrac{1}{90}\)

Answer 1

Đề bài là gì vậy bạn

Answer 2

\(p=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\)

 \(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

Answer 3

\(P=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+....+\dfrac{1}{90}\)

    =\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

    =\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

    =\(\dfrac{1}{1}\)-\(\dfrac{1}{10}\)=\(\dfrac{10}{10}\)-\(\dfrac{1}{10}=\dfrac{9}{10}\)