\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{9.10}\)< 1
Ta có:
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{9.10}\)
= 1 - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+...+\(\dfrac{1}{9}\) - \(\dfrac{1}{10}\)
=1 - \(\dfrac{1}{10}\)
=\(\dfrac{9}{10}\)
=>\(\dfrac{9}{10}\) < 1
Vậy \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{9.10}\) < 1
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.....+\(\dfrac{1}{99.100}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{19.20}\)
B=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
P=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.....+\(\dfrac{1}{99.100}\)
G= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{2003.2004}\)
Tính
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
Tính tổng
$\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$ + $\dfrac{1}{3.4}$ + .... + $\dfrac{1}{99.100}$
tính nhanh
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}=?????\)
