利用有理數加法的交換律和結合律，將下列各式化簡為有理數
(i) \( \frac{2}{5}+\frac{7}{3}+\frac{-4}{5}+\frac{-1}{3} \)
(ii) \( \frac{3}{7}+\frac{-4}{9}+\frac{-11}{7}+\frac{7}{9} \)
(iii) \( \frac{2}{5}+\frac{8}{3}+\frac{-11}{15}+\frac{4}{5}+\frac{-2}{3} \)
(iv) \( \frac{4}{7}+0+\frac{-8}{9}+\frac{-13}{7}+\frac{17}{21} \)
解題步驟
我們需要將每個給定的和表示為有理數。
解答
(i) $\frac{2}{5}+\frac{7}{3}+\frac{-4}{5}+\frac{-1}{3}=(\frac{2}{5}+\frac{-4}{5})+(\frac{7}{3}+\frac{-1}{3})$
$=(\frac{2-4}{5})+(\frac{7-1}{3})$
$=\frac{-2}{5}+\frac{6}{3}$
$=\frac{-2}{5}+\frac{2}{1}$
$=\frac{2}{1}-\frac{2}{5}$
$=\frac{10-2}{5}$
$=\frac{8}{5}$
(ii) $\frac{3}{7}+\frac{-4}{9}+\frac{-11}{7}+\frac{7}{9}=(\frac{3}{7}+\frac{-11}{7})+(\frac{-4}{9}+\frac{7}{9})$
$=\frac{3-11}{7}+\frac{-4+7}{9}$
$=\frac{-8}{7}+\frac{3}{9}$
$=\frac{-8}{7}+\frac{1}{3}$
$=\frac{-24+7}{21}$
$=\frac{-17}{21}$
(iii) $\frac{2}{5}+\frac{8}{3}+\frac{-11}{15}+\frac{4}{5}+\frac{-2}{3}=(\frac{2}{5}+\frac{4}{5})+(\frac{8}{3}+\frac{-2}{3})+\frac{-11}{15}$
$=\frac{2+4}{5}+\frac{8-2}{3}+\frac{-11}{15}$
$=\frac{6}{5}+\frac{6}{3}+\frac{-11}{15}$
$=\frac{18+30-11}{15}$
$=\frac{48-11}{15}$
$=\frac{37}{15}$
(iv) $\frac{4}{7}+0+\frac{-8}{9}+\frac{-13}{7}+\frac{17}{21}=(\frac{4}{7}+\frac{-13}{7})+(\frac{-8}{9}+\frac{17}{21})+0$
$=\frac{4-13}{7}+\frac{-56+51}{63}$
$=\frac{-9}{7}+\frac{-5}{63}$
$=\frac{-81-5}{63}$
$=\frac{-86}{63}$
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