Question

[TL]

Tìm x:

\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

Giúp mình nhanh nhé, cảm ơn các bạn rất nhiều!

Answer 1

[thiên thần mặt trời]

\(\frac{2}{2.3}\) +   \(\frac{2}{3.4}\) +  \(\frac{2}{4.5}\) + .......+ \(\frac{2}{x.\left(x+1\right)}\) = \(\frac{2017}{2019}\) 

2 . (  \(\frac{1}{2}\) -  \(\frac{1}{3}\) + \(\frac{1}{3}\) -  \(\frac{1}{4}\) + .......+  \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)

2 . ( \(\frac{1}{2}\) -  \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)

\(\frac{1}{2}\) -  \(\frac{1}{x+1}\) =  \(\frac{2017}{2019}\) : 2 

 \(\frac{1}{2}\) -  \(\frac{1}{x+1}\) = \(\frac{2017}{4038}\)

             \(\frac{1}{x+1}\)  =  \(\frac{1}{2}\)  -    \(\frac{2017}{4038}\)

              \(\frac{1}{x+1}\)  = \(\frac{1}{2019}\) 

     <=> x + 1 = 2019 => x = 2018

vậy x = 2018

Answer 2

[_Guiltykamikk_]

\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2019}\)

\(\Rightarrow x+1=2019\)

\(\Leftrightarrow x=2018\)

Vậy  \(x=2018\)