Question

Tìm số tự nhiên x biết:

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2009}{2011}\)

Giúp mk nha làm ơn 

Answer 1

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

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2011}:2\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2011}\)

\(\Leftrightarrow x+1=2011\)

\(\Leftrightarrow x=2010\)

Answer 2

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2009}{2011}\)

\(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.......+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(1-\frac{2}{x+1}=\frac{2009}{2011}\)

\(\frac{2}{x+1}=1-\frac{2009}{2011}\)

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

\(x+1=2011\)

\(x=2011-1\)

\(\Rightarrow x=2010\)