Lời giải:
Ta có:
\(\frac{1}{x(x+1):2}=\frac{2}{x(x+1)}=2.\frac{(x+1)-x}{x(x+1)}=2\left(\frac{1}{x}-\frac{1}{x+1}\right)\)
Do đó:
\(\frac{1}{3}=\frac{1}{2.3:2}=2\left(\frac{1}{2}-\frac{1}{3}\right)\)
\(\frac{1}{6}=\frac{1}{3.4:2}=2\left(\frac{1}{3}-\frac{1}{4}\right)\)
\(\frac{1}{10}=\frac{1}{4.5:2}=2\left(\frac{1}{4}-\frac{1}{5}\right)\)
.......
\(\frac{1}{x(x+1):2}=2\left(\frac{1}{x}-\frac{1}{x+1}\right)\)
Cộng theo vế:
\(\text{VT}=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{x}-\frac{1}{x+1}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{x+1}\right)\) \(=1-\frac{2}{x+1}\)
Mà \(\text{VT}=\frac{2009}{2011}\Rightarrow 1-\frac{2}{x+1}=\frac{2009}{2011}\Rightarrow x=2010\)
Mình có cách giải khác:
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right):2}=\dfrac{2009}{2011}\)
=\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2009}{4022}\)
=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2009}{4022}\)
=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2009}{4022}_{ }\)
=\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2009}{4022}\)
➩ \(\dfrac{1}{x+1}=\dfrac{1}{2011}\)
➩ \(x=2011-1=2010\)
