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Hướng Dẫn Giải

\(100 - (1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{99}} + \frac{1}{{100}})\)

\( = \underbrace {(1 + 1 + 1 + ... + 1 + 1)}_{} - (1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{99}} + \frac{1}{{100}})\)

100 số 1

\( = (1 - 1) + (1 - \frac{1}{2}) + (1 - \frac{1}{3}) + ... + (1 - \frac{1}{{99}}) + (1 - \frac{1}{{100}})\)

\( = 0 + \frac{1}{2} + \frac{2}{3} + ... + \frac{{98}}{{99}} + \frac{{99}}{{100}}\)

\( = \frac{1}{2} + \frac{2}{3} + ... + \frac{{98}}{{99}} + \frac{{99}}{{100}}\)

Vậy: \(100 - (1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{99}} + \frac{1}{{100}}) = (\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + ... + \frac{{99}}{{100}})\)