Đáp án: $B>\dfrac12$
Giải thích các bước giải:
Ta có:
$B=(1-\dfrac{1}{2^2})(1-\dfrac{1}{3^2})(1-\dfrac1{4^2})...(1-\dfrac{1}{2021^2})$
$\to B=\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}.\dfrac{4^2-1}{4^2}...\dfrac{2021^2-1}{2021^2}$
$\to B=\dfrac{(2-1)(2+1)}{2^2}.\dfrac{(3-1)(3+1)}{3^2}.\dfrac{(4-1)(4+1)}{4^2}...\dfrac{(2021-1)(2021+1)}{2021^2}$
$\to B=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{2020.2022}{2021^2}$
$\to B=\dfrac{1.2.3...2020}{2.3.4..2021}.\dfrac{3.4.5..2022}{2.3.4...2021}$
$\to B=\dfrac1{2021}.\dfrac{2022}{2}$
$\to B=\dfrac{2022}{2021}.\dfrac{1}{2}>1.\dfrac12$
$\to B>\dfrac12$
