Ta có:
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(\Rightarrow\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2005.2006.2007}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2005.2006}-\frac{1}{2006.2007}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2006.2007}\right)=\frac{1}{2}\left(\frac{2005.2008}{2.2006.2007}\right)\)
Đặt \(A=1.2+2.3+...+n\left(n+1\right)\)
\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+...+n\left(n+1\right)\left(n+2-\left(n-1\right)\right)\)
\(\Rightarrow3A=1.2.3-1.2.0+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(\Rightarrow3A=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
\(\Rightarrow1.2+2.3+...+2006.2007=\frac{2006.2007.2008}{2}\)
Vậy pt trở thành:
\(\frac{1}{2}\left(\frac{2005.2008}{2.2006.2007}\right)x=\frac{2006.2007.2008}{2}\)
\(\Leftrightarrow\frac{2005}{2.2006.2007}x=2006.2007\)
\(\Rightarrow x=\frac{2.\left(2006.2007\right)^2}{2005}\)
