Question

[Trần Tích Thường]

Tìm x :

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\times\left(x+1\right)}=\frac{2011}{2013}\)

Answer 1

[Arima Kousei]

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Leftrightarrow1-\frac{2}{x+1}=\frac{2011}{2013}\)

\(\Leftrightarrow\frac{2}{x+1}=\frac{2}{2013}\)

\(\Leftrightarrow x+1=2013\)

\(\Leftrightarrow x=2012\)

Vậy \(x=2012\)

Answer 2

[Huy Hoang]

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.........+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2005}\left(1\right)\)

\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{x\left(x+1\right)}\)

\(=2.\left[\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{x\left(x+1\right)}\right]\)

\(=2.\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.......+\frac{1}{x}-\frac{1}{x+1}\right]\)

\(=2.\left(1-\frac{1}{x+1}\right)\)

\(=2.\left(\frac{x+1}{x+1}-\frac{1}{x+1}\right)\)

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

Thay vào ( 1 ) ta có :

\(\frac{2x}{x+1}=\frac{4008}{2005}\Rightarrow\frac{x}{x+1}=\frac{2004}{2005}\)

\(\Rightarrow2005x=2004\left(x+1\right)\Rightarrow2005x=2004.2004\)

\(\Rightarrow2005x=2004x=2004x\Rightarrow x=2004\)

KL : Vậy x = 2004

Đây là bài mẫu của mình bạn dựa theo rồi tự làm nhé