Question

giúp mình giải hai bài này nha các bạn ............... thanh kiu <3<3

 

 

 

Answer 1

Câu 9 :

\(A=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{99.100}{2}}+\frac{1}{50}\)

   \(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}+\frac{1}{50}\)

   \(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)

   \(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2.\frac{49}{100}+\frac{1}{50}=\frac{49}{50}+\frac{1}{50}=\frac{50}{50}=1\)