Đáp án:
$ A<-\dfrac{1}{2}$
Giải thích các bước giải:
$A=(\dfrac{1}{2^2}-1)(\dfrac{1}{3^2}-1)(\dfrac{1}{4^2}-1)...(\dfrac{1}{100^2}-1)\\ \rightarrow A=\dfrac{1-2^2}{2^2}.\dfrac{1-3^2}{3^2}...\dfrac{1-100^2}{100^2}\\ \rightarrow A=-\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}...\dfrac{100^2-1}{100^2}\\ \rightarrow A=-\dfrac{(2-1)(2+1)}{2^2}.\dfrac{(3-1)(3+1)}{3^2}...\dfrac{(100-1)(100+1)}{100^2}\\ \rightarrow A=-\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}....\dfrac{99.101}{100.100}\\ \rightarrow A=-\dfrac{1.2.3...99}{2.3.4..100}.\dfrac{3.4...101}{2.3.4...100}\\ \rightarrow A=-\dfrac{1}{100}.\dfrac{101}{2}\\ \rightarrow A=-\dfrac{101}{100.2}<-\dfrac{1}{2}$
