Đáp án:
Giải thích các bước giải:
$\dfrac{200 - \bigg(3 + \dfrac{2}{3} + ... + \dfrac{2}{100}\bigg)}{\dfrac{1}{2} + \dfrac{2}{3} + ... + \dfrac{99}{100}}$
$=\dfrac{200-2-\bigg(\dfrac{2}{2}+\dfrac{2}{3}+...+\dfrac{2}{100}\bigg)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+...+1-\dfrac{1}{100}}$
$=\dfrac{198-\bigg(\dfrac{2}{2}+\dfrac{2}{3}+...+\dfrac{2}{100}\bigg)}{\bigg(1+1+...+1)-(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\bigg)}$
$=\dfrac{2.\bigg[99-\bigg(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\bigg)\bigg]}{99-\bigg(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\bigg)}$
$=2$
