Đáp án:
Giải thích các bước giải:
$1+\frac{2}{6} + \frac{2}{12} + \frac{2}{20} + ... + \frac{2}{x(x+1)}= \frac{1989}{1991}$
$⇒ 2. ( \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + ... + \frac{1}{x(x+1)}) = \frac{1989}{1991} - 1 $
$⇒ \frac{1}{2.3} + \frac{1}{3.4} + \frac{1}{4.5} + ... + \frac{1}{x( x+1) } = \frac{\frac{1989}{1991}-1 }{2} $
$⇒ \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + ... + \frac{1}{x} - \frac{1}{x+1} = \frac{-1}{1991} $
$⇒ \frac{1}{2} - \frac{1}{x+1} = \frac{-1}{1991} $
$⇒ \frac{1}{x+1} = \frac{1}{2} - ( \frac{-1}{1991} ) $
$⇒ \frac{1}{x+1} = \frac{1993}{3982} $
$⇒ x + 1 = \frac{3982}{1993} $
$⇒ x = \frac{3982}{1993} - 1 $
$⇒ x = \frac{1989}{1993} $
$ \text{ Vậy } x = \frac{1989}{1993} $
