Đáp án:
a) `2001`
b) `127/128`
Giải thích các bước giải:
a) $6 \dfrac{2}{7}+7 \dfrac{3}{5}+8 \dfrac{6}{9}+9 \dfrac{1}{4}+\dfrac{2}{5}+\dfrac{5}{7}+\dfrac{1}{3}+\dfrac{3}{4}+1967\\= \dfrac{44}{7}+\dfrac{38}{5}+ \dfrac{26}{3}+ \dfrac{37}{4}+ \dfrac{2}{5}+\dfrac{5}{7}+\dfrac{1}{3}+\dfrac{3}{4}+1967\\= \bigg(\dfrac{44}{7}+\dfrac{5}{7}\bigg)+\bigg(\dfrac{38}{5}+\dfrac{2}{5}\bigg)+\bigg(\dfrac{26}{3}+\dfrac{1}{3}+\dfrac{37}{4}+\dfrac{3}{4}\bigg)+1967\\=7+8+9+10+1967=2001$
b) $\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\\= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{128}\\=1 - \dfrac{1}{128}=\dfrac{127}{128}$
