Đáp án:
$A>\dfrac{3}{4}$
Giải thích các bước giải:
$A= \dfrac{1}{2\times 2}+ \dfrac{1}{3\times 3}+ \dfrac{1}{4\times 4}+...+ \dfrac{1}{2020\times 2020}$
Ta thấy:
$ \dfrac{1}{2\times 2}> \dfrac{1}{2\times 3}$
$ \dfrac{1}{3\times 3}> \dfrac{1}{3\times 4}$
$ \dfrac{1}{4\times 4}> \dfrac{1}{4\times 5}$
.......
$ \dfrac{1}{2020\times 2020}> \dfrac{1}{2020\times 2021}$
$\Rightarrow A> \dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+\dfrac{1}{4\times 5}+...+\dfrac{1}{2020\times 2021}$
$\Rightarrow A> \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+.....+\dfrac{1}{2020}-\dfrac{1}{2021}$
$\Rightarrow A> 1-\dfrac{1}{2021}$
$\Rightarrow A<\dfrac{2020}{2021}$
Mặt khác: $\dfrac{2020}{2021}>\dfrac{3}{4}$
$\Rightarrow A>\dfrac{3}{4}$
