1 + 2/6 + 2/ 12+…+ 2/x .(x + 1) = 1989/1991

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1. 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} $                    

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2. Đáp án:

    

   Giải thích các bước giải:

    $1+\dfrac{2}{6}+\dfrac{2}{12}+…+\dfrac{2}{x.(x+1)}=\dfrac{1989}{1991}$

   $ $

   $⇒2.(\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{x.(x+1)})=\dfrac{-2}{1991}$

   $ $

   $⇒2.(\dfrac{1}{2}-\dfrac{1}{x+1})=\dfrac{-2}{1991}$

   $ $ 

   $⇒\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{-1}{1991}$

   $ $

   $⇒\dfrac{1}{x+1}=\dfrac{1993}{3982}$

   $ $

   $⇒\dfrac{1993}{1993x+1993}=\dfrac{1993}{3982}$

   $ $

   $⇒1993x+1993=3982$

   $⇒1993x=1989$

   $⇒x=\dfrac{1989}{1993}$

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