Tính nhanh $\dfrac{1}{1.2.3}$ + $\dfrac{1}{2.3.4}$ + $\dfrac{1}{3.4.5}$+ ... + $\dfrac{1}{10.11.12}$ $\dfrac{1}{19}$ + $\dfrac{9}{19.29}$ + $\dfrac{9}{29.30}$ +..... $\dfrac{9}{1999.2009}$
1 câu trả lời
Đáp án+Giải thích các bước giải:
a. $\frac{1}{1.2.3}$ + $\frac{1}{2.3.4}$ + $\frac{1}{3.4.5}$ +...+ $\frac{1}{10.11.12}$
= $\frac{1}{1.2}$ - $\frac{1}{2.3}$ + $\frac{1}{2.3}$ - $\frac{1}{3.4}$ + $\frac{1}{3.4}$ - $\frac{1}{4.5}$ +...+ $\frac{1}{10.11}$ - $\frac{1}{11.12}$
= $\frac{1}{1.2}$ - $\frac{1}{11.12}$ = $\frac{1}{2}$ - $\frac{1}{132}$ = $\frac{65}{132}$
b. $\frac{1}{19}$ + $\frac{9}{19.29}$ + $\frac{9}{29.39}$ +...+ $\frac{9}{1999.2009}$
= $\frac{9}{9.19}$ + $\frac{9}{19.29}$ + $\frac{9}{29.39}$ +...+ $\frac{9}{1999.2009}$
= $\frac{9}{10}$.($\frac{19-9}{9.19}$+ $\frac{29-19}{19.29}$+ $\frac{39-29}{29.39}$ +...+ $\frac{2009-1999}{1999.2009}$)
= $\frac{9}{10}$.($\frac{1}{9}$- $\frac{1}{19}$+ $\frac{1}{19}$ -$\frac{1}{29}$ +$\frac{1}{29}$ - $\frac{1}{39}$+...+ $\frac{1}{1999}$ - $\frac{1}{2009}$)
= $\frac{9}{10}$ .($\frac{1}{9}$- $\frac{1}{2009}$)
= $\frac{1}{10}$ - $\frac{9}{2009.10}$
= $\frac{2009}{2009.10}$ - $\frac{9}{2009.10}$
= $\frac{2000}{2009.10}$
= $\frac{200}{2009}$
