Question

tim x : 1+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) +..............+\(\dfrac{1}{x\left(x+2\right):2}\) =1\(\dfrac{2009}{2011}\)

Answer 1

\(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+2\right)}=1\dfrac{2009}{2011}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{4020}{4022}\)

\(\Rightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{4020}{4022}\)

\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4020}{4022}\)

\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{4020}{4022}\)

\(\Rightarrow\dfrac{1}{x+2}=\dfrac{1}{2011}\)

\(\Rightarrow x+2=2011\Rightarrow x=2009\)

Vậy x = 2009