Sửa đề :
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x}+\frac{1}{x\left(x+1\right)}=\frac{504}{1009}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2018}\)
\(\Leftrightarrow x=2018-1\)
\(\Leftrightarrow x=2017\)
Vậy ...
Sửa đề \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2018}\)
\(\Leftrightarrow x=2017\)
