Question

a)\(\dfrac{3}{5x8}+\dfrac{3}{8x11}+...+\dfrac{3}{2006x2009}\)

b)\(\dfrac{1}{6x10}+\dfrac{1}{10x14}+...+\dfrac{1}{402x406}\)

c)\(\dfrac{10}{7x12}+\dfrac{10}{12x17}+...+\dfrac{10}{502x507}\)

Chú ý: x ở đề bài là dấu nhân nha các bạn.Mong các bạn giúp mk nhanh 1 chút vì mk đag cần gấp

Answer 1

a)

\(A=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2006.2009}\)

\(=\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+....+\frac{2009-2006}{2006.2009}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)

\(=\frac{1}{5}-\frac{1}{2009}=\frac{2004}{10045}\)

Answer 2

b)

\(B=\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{402.406}\)

\(\Rightarrow 4B=\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{402.406}\)

\(4B=\frac{10-6}{6.10}+\frac{14-10}{10.14}+...+\frac{406-402}{402.406}\)

\(4B=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{402}-\frac{1}{406}\)

\(4B=\frac{1}{6}-\frac{1}{406}=\frac{100}{609}\Rightarrow B=\frac{25}{609}\)

Answer 3

c)

\(C=\frac{10}{7,12}+\frac{10}{12.17}+...+\frac{10}{502.507}\)

\(\Rightarrow \frac{C}{2}=\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{502.507}\)

\(\frac{C}{2}=\frac{12-7}{7.12}+\frac{17-12}{12.17}+...+\frac{507-502}{502.507}\)

\(\frac{C}{2}=\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+....+\frac{1}{502}-\frac{1}{507}\)

\(\frac{C}{2}=\frac{1}{7}-\frac{1}{507}=\frac{500}{3549}\)

\(\Rightarrow C=\frac{1000}{3549}\)