Đặt A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{2013}{2014}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2014}\)
\(\Rightarrow A=1-\frac{1}{x+1}=\frac{2013}{2014}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2013}{2014}\)
\(\Rightarrow\)\(\frac{1}{x+1}=\frac{1}{2014}\)
\(\Rightarrow x+1=2014\)
\(\Rightarrow x=2014-1\)
\(\Rightarrow x=2013\)
Vậy x=2013
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2014}\)
\(1-\frac{1}{x+1}=\frac{2013}{2014}\)
\(\frac{1}{x+1}=1-\frac{2013}{2014}\)
\(\frac{1}{x+1}=\frac{1}{2014}\)
Vì \(x+1\)là mẫu số nên:
\(x+1=2014\)
\(x=2014-1=2013\)
Vậy ....
P/s: Dấu . là nhân nha!
