Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.

\(\frac{1}{1}\)x 2 x 3 + \(\frac{1}{2}\)x 3 x 4 + \(\frac{1}{3}\)x 4 x 5 + \(\frac{1}{4}\)x 5 x 6

= 1 x 2 + \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)x 6

= 2 +\(\frac{1}{2}\)+ \(\frac{1}{3}\)+ 1, 5

=

\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}\)

\(2A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{20-18}{18.19.20}=\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}=\dfrac{1}{2}-\dfrac{1}{19.20}\)

\(\Rightarrow A=\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right):2\)

11 + báo cáo

=1/2 - 1/101

Sửa đề :

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(1-\frac{1}{101}=\frac{100}{101}\)

Đặt: \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2011.2013}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2013}\right)\)

\(=\frac{1}{2}.\frac{2012}{2013}\)

\(=\frac{1006}{2013}\)

Em nên gõ công thức trực quan để đề bài rõ ràng nhé

Đặt A=\(\frac{1}{3}.5+\frac{1}{5}.7+...+\frac{1}{97}.99\)

=>A=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

=>2A=\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

=>2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

=>2A=\(\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)

=>A=\(\frac{32}{99}:2=\frac{32}{99}.\frac{1}{2}=\frac{32}{198}=\frac{16}{99}\)