\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)...\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\left(\dfrac{-1}{2}\right).\dfrac{3}{2}.\left(-\dfrac{2}{3}\right).\dfrac{4}{3}...\left(\dfrac{-99}{100}\right).\dfrac{101}{100}\)

\(=\dfrac{\left(-1\right).\left(-2\right)...\left(-99\right)}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)

\(=\dfrac{1.2...99}{2.3...100}.\dfrac{101}{2}\)

\(=\dfrac{1}{100}.\dfrac{101}{2}\)

\(=\dfrac{101}{200}>\dfrac{100}{200}=\dfrac{1}{2}\)

Vậy...

Nhận thấy A có 99 hạng tử mà mỗi hạng tử chứa dấu âm nên viết gọn\(A=-\dfrac{3}{4}.\dfrac{8}{9}.....\dfrac{9999}{10000}=-\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}....\dfrac{99.101}{100^2}=-\dfrac{\left(1.2...99\right).\left(3.4...101\right)}{\left(2.3..100\right).\left(2.3...100\right)}=-\dfrac{101}{2.100}=-\dfrac{101}{200}< -\dfrac{1}{2}\)

\(A=1+\frac{2^2}{3^2}+\frac{2^2}{5^2}+\frac{2^2}{7^2}+...+\frac{2^2}{2009^2}\)

\(A=1+2^2\left(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+..+\frac{1}{2009^2}\right)\)

Ta có: \(\frac{1}{3^2}< \frac{1}{1.3};\frac{1}{5^2}< \frac{1}{3.5};\frac{1}{7^2}< \frac{1}{5.7};...;\frac{1}{2009^2}< \frac{1}{2007.2009}\)

\(\Rightarrow A< 1+4\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{2007.2009}\right)\)

\(=1+4\cdot\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2007}-\frac{1}{2009}\right)\)

\(=1+2\left(1-\frac{1}{2009}\right)=3-\frac{2}{2009}< 3\)

\(\Rightarrow A< 3\)

biến đổi vế trái ta được:

A=1/1-1/2+1/3-1/4+...+1/99-1/100

  =(1/1+1/2+... +1/100)-2(1/2+1/4+....+1/100)

  =(1/1+1/2+...+1/100)-(1/1+1/2+....+1/50)

  =1/51+1/52+....+1/100

vậy A=B

Đặt \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\)

\(A>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\) ( 19 số hạng )

\(A>\frac{19}{20}\)