Question

chứng tỏ A =\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{3.7}+...+\frac{2}{99.101}\) >1

Answer 1

biểu thức trên = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=\frac{100}{101}< 1\)

vậy A<1

Answer 2

Ta thấy

Answer 3

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

\(=\left(\frac{1}{1}+\frac{1}{101}\right)\)

\(=\frac{102}{101}\)

\(\Rightarrow A>1\)

Answer 4

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{3.7}+...+\frac{2}{99.101}>1\)

\(A=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(A=2.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

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

\(A=\frac{100}{101}\)

\(\Rightarrow A=\frac{100}{101}< 1\left(đpcm\right)\)

Answer 5

cậu ơi hình như sai đề