a,Ta có \(\dfrac{1}{2.3}\)=\(\dfrac{1}{6}\)
\(\dfrac{1}{2}-\dfrac{1}{3}\)=\(\dfrac{3}{6}-\dfrac{2}{6}\)=\(\dfrac{1}{6}\)
=>\(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
b, \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2005.2006}\)
=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{2005}-\dfrac{1}{2006}\)
=\(\dfrac{1}{1}-\dfrac{1}{2006}\)
=\(\dfrac{2006}{2006}-\dfrac{1}{2006}\)
=\(\dfrac{2005}{2006}\)
Ta có
\(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{\left(n+1\right)-n}{n.\left(n+1\right)}=\dfrac{1}{n.\left(n+1\right)}\)
Vậy \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
Áp dụng từ a, ta có
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2005.2006}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2005}-\dfrac{1}{2006}\)
= \(1-\dfrac{1}{2006}\)
= \(\dfrac{2005}{2006}\)
a) Bằng nhau
b) 1/1.2 + 1/2.3 + 1/3.4 + ..... + 1/2005.2006
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2005 - 1/2006
= 1/1 - 1/2006
= 2005/2006
