Đáp án:
Giải thích các bước giải:
đề
$A= \dfrac{1- 2^2}{2^2}. \dfrac{1- 3^2}{3^2}. ....\dfrac{1- 100^2}{100^2}$
$A= -\dfrac{2^2- 1}{2^2}. \dfrac{3^2- 1}{3^2}...\dfrac{100^2- 1}{100^2}$
$A= -\dfrac{(2- 1) (2+ 1)}{2^2}. \dfrac{(3- 1) (3+ 1)}{3^2}... \dfrac{(100- 1) (100+ 1)}{100^2}$
$A= -\dfrac{1. 3}{2. 2}. \dfrac{2. 4}{3. 3}. \dfrac{99. 101}{100. 100}$
$A= -\dfrac{1. 2. 3...99}{2. 3. 4...100}. \dfrac{3. 4... 101}{2. 3. 4...100}$
$A= -\dfrac{1}{100}. \dfrac{101}{2}$
$⇒ A= -\dfrac{101}{100. 2}< -\dfrac{1}{2} (ddpcm)$
