Question

Chứng tỏ rằng:

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{67.69}>\frac{11}{35}\)

Answer 1

Ta có :

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{67.69}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{67}-\frac{1}{69}\)

\(=\frac{1}{3}-\frac{1}{69}\)

\(=\frac{22}{69}\)

Mà \(\frac{22}{69}>\frac{11}{35}\left(=\frac{22}{70}\right)\)

\(\Rightarrowđpcm\)

Answer 2

Ta có: \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{67.69}\)

= \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{69}\)

= \(\frac{1}{3}-\frac{1}{69}\)

= \(\frac{22}{69}\)> 11/35

Answer 3

Có:\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{67.69}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{69}\)

       \(=\frac{1}{3}-\frac{1}{69}=\frac{22}{69}\left(1\right)\)

    \(\frac{11}{35}=\frac{22}{70}\left(2\right)\)

Từ (1)và (2)=>đpcm