Đáp án+Giải thích các bước giải:
Bài 9 :
2) $\dfrac{2}{1.3}$ + $\dfrac{2}{3.5}$ + $\dfrac{2}{5.7}$ + ... + $\dfrac{2}{2003.2005}$
= $\dfrac{1}{1}$ - $\dfrac{1}{3}$ + $\dfrac{1}{3}$ - $\dfrac{1}{5}$ + $\dfrac{1}{5}$ - $\dfrac{1}{7}$ + ... +$\dfrac{1}{2003}$ - $\dfrac{1}{2005}$
= 1 - $\dfrac{1}{2005}$
= $\dfrac{2004}{2005}$
4) $\dfrac{3}{4.7}$ + $\dfrac{3}{7.10}$ + $\dfrac{3}{10.13}$ + ... + $\dfrac{3}{97.100}$
= $\dfrac{1}{4}$ - $\dfrac{1}{7}$ + $\dfrac{1}{7}$ - $\dfrac{1}{10}$ + $\dfrac{1}{10}$ - $\dfrac{1}{13}$ + ... +$\dfrac{1}{97}$ - $\dfrac{1}{100}$
= $\dfrac{1}{4}$ - $\dfrac{1}{100}$
= $\dfrac{24}{100}$
= $\dfrac{6}{25}$
Bài 11 :
1) ý đầu tiên này có phải bạn viết nhầm không
2) ($\dfrac{1}{2.3}$ + $\dfrac{1}{3.4}$ + ... + $\dfrac{1}{49.50}$) . x = 1
⇒ ($\dfrac{1}{2}$ - $\dfrac{1}{3}$ + $\dfrac{1}{3}$ - $\dfrac{1}{4}$ + ... + $\dfrac{1}{49}$ - $\dfrac{1}{50}$ ) . x = 1
⇒ ($\dfrac{1}{2}$ - $\dfrac{1}{50}$) . x = 1
⇒ $\dfrac{24}{50}$ . x = 1
⇒ x = $\dfrac{25}{12}$
