Chương III : Phân số

Tanya

Tìm x:

\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{2015}{2017}\)

Nguyễn Trần Diệu Linh
8 tháng 4 2018 lúc 10:01

1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + ... + \(\dfrac{2}{x\left(x+1\right)}\) = 1\(\dfrac{2015}{2017}\)

=> \(\dfrac{2}{2}\) + \(\dfrac{2}{6}\) + \(\dfrac{2}{12}\) + \(\dfrac{2}{20}\) + ... + \(\dfrac{2}{x\left(x+1\right)}\) = 1\(\dfrac{2015}{2017}\)

=> \(\dfrac{2}{1.2}\) + \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{x\left(x+1\right)}\) = 1\(\dfrac{2015}{2017}\)

=> \(\dfrac{1.2}{1.2}\) + \(\dfrac{1.2}{2.3}\) + \(\dfrac{1.2}{3.4}\) + \(\dfrac{1.2}{4.5}\) + ... + \(\dfrac{1.2}{x\left(x+1\right)}\) = 1\(\dfrac{2015}{2017}\)

=> 2(\(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{x\left(x+1\right)}\)) = 1\(\dfrac{2015}{2017}\)

=> 2(1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\)) = 1\(\dfrac{2015}{2017}\)

=> 2(1 - \(\dfrac{1}{x+1}\)) = \(\dfrac{4032}{2017}\)

=> 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{4032}{2017}\) : 2

=> 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{2016}{2017}\)

=> \(\dfrac{1}{x+1}\) = 1 - \(\dfrac{2016}{2017}\)

=> \(\dfrac{1}{x+1}\) = \(\dfrac{1}{2017}\)

=> x + 1 = 2017

=> x = 2017 - 1

=> x = 2016

Bình luận (2)

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