問１ 分母が同じ分数のたし算・ひき算
\(（１）\dfrac{2}{7}\ +\ \dfrac{3}{7}\ =\)
- 答え（１）
- \(\dfrac{5}{7}\)
- 解き方（１）
- \(\dfrac{2}{7}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{7}\)
\(（２）\dfrac{8}{9}\ -\ \dfrac{1}{9}\ =\)
- 答え（２）
- \(\dfrac{7}{9}\)
- 解き方（２）
- \(\dfrac{8}{9}\) – \(\dfrac{1}{9}\) = \(\dfrac{7}{9}\)
\(（３）\dfrac{7}{9}\ +\ \dfrac{4}{9}\ =\)
- 答え（３）
- \(1\dfrac{2}{9}\)
- 解き方（３）
- \(\dfrac{7}{9}\) + \(\dfrac{4}{9}\) = \(\dfrac{11}{9}\) = \(1\dfrac{2}{9}\)
※答えは帯分数に直す
\(（４）1\ -\ \dfrac{1}{9}\ =\)
- 答え（４）
- \(\dfrac{8}{9}\)
- 解き方（４）
- 1 – \(\dfrac{1}{9}\) = \(\dfrac{9}{9}\) – \(\dfrac{1}{9}\) = \(\dfrac{8}{9}\)
\(（５）6\dfrac{5}{13}\ +\ 4\dfrac{6}{13}\ =\)
- 答え（５）
- \(10\dfrac{11}{13}\)
- 解き方（５）
- \(6\dfrac{5}{13}\) + \(4\dfrac{6}{13}\) = (6 + 4) + ( \(\dfrac{5}{13}\) + \(\dfrac{6}{13}\) ) = \(10\dfrac{11}{13}\)
\(（６）8\dfrac{2}{13}\ -\ 7\dfrac{4}{13}\ =\)
- 答え（６）
- \(\dfrac{11}{13}\)
- 解き方（６）
- \(8\dfrac{2}{13}\) – \(7\dfrac{4}{13}\) = \(7\dfrac{15}{13}\) – \(7\dfrac{4}{13}\) = (7 – 7) + ( \(\dfrac{15}{13}\) – \(\dfrac{4}{13}\) ) = \(\dfrac{11}{13}\)
\(（７）\dfrac{8}{15}\ -\ \dfrac{2}{15}\ +\ \dfrac{4}{15}\ =\)
- 答え（７）
- \(\dfrac{2}{3}\)
- 解き方（７）
- \(\dfrac{8}{15}\) – \(\dfrac{2}{15}\) + \(\dfrac{4}{15}\) = \(\dfrac{10}{15}\) = \(\dfrac{2}{3}\)
※答えは既約分数に直す
\(（８）12\ -\ 9\dfrac{1}{8}\ +\ 3\dfrac{5}{8}\ =\)
- 答え（８）
- \(6\dfrac{1}{2}\)
- 解き方（８）
- 12 – \(9\dfrac{1}{8}\) + \(3\dfrac{5}{8}\)
= \(11\dfrac{8}{8}\) – \(9\dfrac{1}{8}\) + \(3\dfrac{5}{8}\)
= {(11 – 9 + 3) + ( \(\dfrac{8}{8}\) – \(\dfrac{1}{8}\) + \(\dfrac{5}{8}\) )}
= \(5\dfrac{12}{8}\) = \(5\dfrac{3}{2}\) = \(6\dfrac{1}{2}\)
\(（９）10\dfrac{1}{9}\ -\ (\ 3\dfrac{2}{9}\ +\ 4\dfrac{8}{9}\ )\ +\ 1\dfrac{4}{9}\ =\)
- 答え（９）
- \(3\dfrac{4}{9}\)
- 解き方（９）
- \(10\dfrac{1}{9}\) – ( \(3\dfrac{2}{9}\) + \(4\dfrac{8}{9}\) ) + \(1\dfrac{4}{9}\)
= \(10\dfrac{1}{9}\) – {(3 + 4) + ( \(\dfrac{2}{9}\) + \(\dfrac{8}{9}\) )} + \(1\dfrac{4}{9}\)
= \(10\dfrac{1}{9}\) – \(7\dfrac{10}{9}\) + \(1\dfrac{4}{9}\)
= \(10\dfrac{1}{9}\) – \(8\dfrac{1}{9}\) + \(1\dfrac{4}{9}\)
= (10 – 8 + 1) + ( \(\dfrac{1}{9}\) – \(\dfrac{1}{9}\) + \(\dfrac{4}{9}\) ) = \(3\dfrac{4}{9}\)
\(（１０）12\ -\ \{\ 5\dfrac{3}{17}\ -\ (\ 2\dfrac{8}{17}\ +\ \dfrac{15}{17}\ )\ \}\ -\ 3\dfrac{9}{17}\ =\)
- 答え（１０）
- \(6\dfrac{11}{17}\)
- 解き方（１０）
- 12 – { \(5\dfrac{3}{17}\) – ( \(2\dfrac{8}{17}\) + \(\dfrac{15}{17}\) )} – \(3\dfrac{9}{17}\)
= 12 – ( \(5\dfrac{3}{17}\) – \(3\dfrac{6}{17}\) ) – \(3\dfrac{9}{17}\)
= 12 – \(1\dfrac{14}{17}\) – \(3\dfrac{9}{17}\)
= \(6\dfrac{11}{17}\)
問２ 分母が異なる分数のたし算・ひき算
\(（１）\dfrac{2}{3}\ +\ \dfrac{1}{6}\ =\)
- 答え（１）
- \(\dfrac{5}{6}\)
- 解き方（１）
- \(\dfrac{2}{3}\) + \(\dfrac{1}{6}\) = \(\dfrac{4}{6}\) + \(\dfrac{1}{6}\) = \(\dfrac{5}{6}\)
\(（２）2\dfrac{11}{27}\ +\ 5\dfrac{7}{9}\ =\)
- 答え（２）
- \(8\dfrac{5}{27}\)
- 解き方（２）
- \(2\dfrac{11}{27}\) + \(5\dfrac{7}{9}\)
= \(2\dfrac{11}{27}\) + \(5\dfrac{21}{27}\) = \(7\dfrac{32}{27}\) = \(8\dfrac{5}{27}\)
\(（３）\dfrac{4}{5}\ -\ \dfrac{1}{7}\ =\)
- 答え（３）
- \(\dfrac{23}{35}\)
- 解き方（３）
- \(\dfrac{4}{5}\) – \(\dfrac{1}{7}\) = \(\dfrac{28}{35}\) – \(\dfrac{5}{35}\) = \(\dfrac{23}{35}\)
\(（４）7\dfrac{1}{6}\ -\ 3\dfrac{7}{9}\ =\)
- 答え（４）
- \(3\dfrac{7}{18}\)
- 解き方（４）
- \(7\dfrac{1}{6}\) – \(3\dfrac{7}{9}\)
= \(7\dfrac{3}{18}\) – \(3\dfrac{14}{18}\)
= \(6\dfrac{21}{18}\) – \(3\dfrac{14}{18}\) = \(3\dfrac{7}{18}\)
\(（５）\dfrac{3}{4}\ +\ \dfrac{2}{3}\ +\ \dfrac{3}{10}\ =\)
- 答え（５）
- \(1\dfrac{43}{60}\)
- 解き方（５）
- \(\dfrac{3}{4}\) + \(\dfrac{2}{3}\) + \(\dfrac{3}{10}\)
= \(\dfrac{9}{12}\) + \(\dfrac{8}{12}\) + \(\dfrac{3}{10}\)
= \(\dfrac{17}{12}\) + \(\dfrac{3}{10}\)
= \(\dfrac{85}{60}\) + \(\dfrac{18}{60}\) = \(\dfrac{103}{60}\) = \(1\dfrac{43}{60}\)
\(（６）15\ -\ \dfrac{5}{16}\ +\ \dfrac{3}{4}\ =\)
- 答え（６）
- \(15\dfrac{7}{16}\)
- 解き方（６）
- 15 – \(\dfrac{5}{16}\) + \(\dfrac{3}{4}\)
= \(14\dfrac{16}{16}\) – \(\dfrac{5}{16}\) + \(\dfrac{12}{16}\)
= \(14\dfrac{23}{16}\) = \(15\dfrac{7}{16}\)
\(（７）5\dfrac{1}{4}\ +\ 4\dfrac{5}{6}\ -\ 3\dfrac{3}{8}\ =\)
- 答え（７）
- \(6\dfrac{17}{24}\)
- 解き方（７）
- \(5\dfrac{1}{4}\) + \(4\dfrac{5}{6}\) – \(3\dfrac{3}{8}\)
= \(5\dfrac{3}{12}\) + \(4\dfrac{10}{12}\) – \(3\dfrac{3}{8}\)
= \(9\dfrac{13}{12}\) – \(3\dfrac{3}{8}\)
= \(9\dfrac{26}{24}\) – \(3\dfrac{9}{24}\) = \(6\dfrac{17}{24}\)
\(（８）5\dfrac{9}{10}\ +\ 4\dfrac{1}{2}\ -\ 3\dfrac{2}{5}\ =\)
- 答え（８）
- 7
- 解き方（８）
- \(5\dfrac{9}{10}\) + \(4\dfrac{1}{2}\) – \(3\dfrac{2}{5}\)
= \(5\dfrac{9}{10}\) + \(4\dfrac{5}{10}\) – \(3\dfrac{2}{5}\)
= \(9\dfrac{14}{10}\) – \(3\dfrac{2}{5}\)
= \(9\dfrac{7}{5}\) – \(3\dfrac{2}{5}\) = \(6\dfrac{5}{5}\) = 7
\(（９）10\dfrac{1}{5}\ -\ (\ 2\dfrac{2}{5}\ +\ 2\dfrac{3}{10}\ )\ +\ 3\dfrac{1}{2}\ =\)
- 答え（９）
- 9
- 解き方（９）
- \(10\dfrac{1}{5}\) – ( \(2\dfrac{2}{5}\) + \(2\dfrac{3}{10}\) ) + \(3\dfrac{1}{2}\)
= \(10\dfrac{1}{5}\) – ( \(2\dfrac{4}{10}\) + \(2\dfrac{3}{10}\) ) + \(3\dfrac{1}{2}\)
= \(10\dfrac{1}{5}\) – \(4\dfrac{7}{10}\) + \(3\dfrac{1}{2}\)
= \(10\dfrac{2}{10}\) – \(4\dfrac{7}{10}\) + \(3\dfrac{1}{2}\)
= \(9\dfrac{12}{10}\) – \(4\dfrac{7}{10}\) + \(3\dfrac{1}{2}\)
= \(5\dfrac{5}{10}\) + \(3\dfrac{1}{2}\)
= \(5\dfrac{1}{2}\) + \(3\dfrac{1}{2}\) = \(8\dfrac{2}{2}\) = 9
\(（１０）50\ -\ \{\ 12\dfrac{1}{6}\ -\ (\ 3\dfrac{3}{4}\ +\ 4\dfrac{5}{8}\ )\ \}\ -\ \dfrac{1}{6}\ =\)
- 答え（１０）
- \(46\dfrac{1}{24}\)
- 解き方（１０）
- 50 – { \(12\dfrac{1}{6}\) – ( \(3\dfrac{3}{4}\) + \(4\dfrac{5}{8}\) )} – \(\dfrac{1}{6}\)
= 50 – { \(12\dfrac{1}{6}\) – ( \(3\dfrac{6}{8}\) + \(4\dfrac{5}{8}\) )} – \(\dfrac{1}{6}\)
= 50 – ( \(12\dfrac{1}{6}\) – \(7\dfrac{11}{8}\) ) – \(\dfrac{1}{6}\)
= 50 – ( \(12\dfrac{4}{24}\) – \(7\dfrac{33}{24}\) ) – \(\dfrac{1}{6}\)
= 50 – ( \(11\dfrac{28}{24}\) – \(7\dfrac{33}{24}\) ) – \(\dfrac{1}{6}\)
= 50 – ( \(11\dfrac{28}{24}\) – \(8\dfrac{9}{24}\) ) – \(\dfrac{1}{6}\)
= 50 – \(3\dfrac{19}{24}\) – \(\dfrac{1}{6}\)
= \(49\dfrac{24}{24}\) – \(3\dfrac{19}{24}\) – \(\dfrac{4}{24}\) = \(46\dfrac{1}{24}\)
問３ かけ算
\(\dfrac{a}{b}\ ×\ \dfrac{c}{d}\ =\ \dfrac{a\ ×\ c}{b\ ×\ d}\)
\(（１）\dfrac{2}{3}\ ×\ \dfrac{4}{5}\ =\)
- 答え（１）
- \(\dfrac{8}{15}\)
- 解き方（１）
- \(\dfrac{2}{3}\) × \(\dfrac{4}{5}\) = \(\dfrac{2\ ×\ 4}{3\ ×\ 5}\) = \(\dfrac{8}{15}\)
\(（２）\dfrac{4}{15}\ ×\ \dfrac{5}{8}\ =\)
- 答え（２）
- \(\dfrac{1}{6}\)
- 解き方（２）
- \(\dfrac{4}{15}\) × \(\dfrac{5}{8}\) = \(\dfrac{\cancelto{1}{4}\ ×\ \cancelto{1}{5}}{\cancelto{3}{15}\ ×\ \cancelto{2}{8}}\) = \(\dfrac{1\ ×\ 1}{3\ ×\ 2}\) = \(\dfrac{1}{6}\)
※かけ算する前に約分する
\(（３）\dfrac{7}{24}\ ×\ 6\ =\)
- 答え（３）
- \(1\dfrac{3}{4}\)
- 解き方（３）
- \(\dfrac{7}{24}\) × 6 = \(\dfrac{7}{24}\) × \(\dfrac{6}{1}\) = \(\dfrac{7\ ×\ \cancelto{1}{6}}{\cancelto{4}{24}\ ×\ 1}\) = \(\dfrac{7\ ×\ 1}{4\ ×\ 1}\) = \(\dfrac{7}{4}\) = \(1\dfrac{3}{4}\)
※かけ算する前に約分する
\(（４）\dfrac{5}{27}\ ×\ \dfrac{9}{17}\ ×\ \dfrac{8}{15}\ =\)
- 答え（４）
- \(\dfrac{8}{153}\)
- 解き方（４）
- \(\dfrac{5}{27}\) × \(\dfrac{9}{17}\) × \(\dfrac{8}{15}\) = \(\dfrac{\cancelto{1}{5}\ ×\ \cancelto{1}{9}\ ×\ 8}{\cancelto{3}{27}\ ×\ 17\ ×\ \cancelto{3}{15}}\) = \(\dfrac{1\ ×\ 1\ ×\ 8}{3\ ×\ 17\ ×\ 3}\) = \(\dfrac{8}{153}\)
※かけ算する前に約分する
\(（５）5\dfrac{1}{3}\ ×\ \dfrac{1}{7}\ =\)
- 答え（５）
- \(\dfrac{16}{21}\)
- 解き方（５）
- \(5\dfrac{1}{3}\) × \(\dfrac{1}{7}\)
= \(\dfrac{16}{3}\) × \(\dfrac{1}{7}\)
※帯分数は仮分数に直す
= \(\dfrac{16\ ×\ 1}{3\ ×\ 7}\) = \(\dfrac{16}{21}\)
\(（６）\dfrac{4}{7}\ ×\ 3\dfrac{1}{9}\ =\)
- 答え（６）
- \(1\dfrac{7}{9}\)
- 解き方（６）
- \(\dfrac{4}{7}\) × \(3\dfrac{1}{9}\)
= \(\dfrac{4}{\cancelto{1}{7}}\) × \(\dfrac{\cancelto{4}{28}}{9}\)
※帯分数は仮分数に直す
※かけ算する前に約分する
= \(\dfrac{4\ ×\ 4}{1\ ×\ 9}\) = \(\dfrac{16}{9}\) = \(1\dfrac{7}{9}\)
※答えは帯分数に直す
\(（７）2\dfrac{2}{9}\ ×\ 1\dfrac{3}{16}\ =\)
- 答え（７）
- \(2\dfrac{23}{36}\)
- 解き方（７）
- \(2\dfrac{2}{9}\) × \(1\dfrac{3}{16}\)
= \(\dfrac{\cancelto{5}{20}}{9}\) × \(\dfrac{19}{\cancelto{4}{16}}\)
※帯分数は仮分数に直す
※かけ算する前に約分する
= \(\dfrac{5\ ×\ 19}{9\ ×\ 4}\) = \(\dfrac{95}{36}\) = \(2\dfrac{23}{36}\)
※答えは帯分数に直す
\(（８）4\dfrac{1}{6}\ ×\ 2\dfrac{7}{10}\ =\)
- 答え（８）
- \(11\dfrac{1}{4}\)
- 解き方（８）
- \(4\dfrac{1}{6}\) × \(2\dfrac{7}{10}\)
= \(\dfrac{\cancelto{5}{25}}{\cancelto{2}{6}}\) × \(\dfrac{\cancelto{9}{27}}{\cancelto{2}{10}}\)
※帯分数は仮分数に直す
※かけ算する前に約分する
= \(\dfrac{5\ ×\ 9}{2\ ×\ 2}\) = \(\dfrac{45}{4}\) = \(11\dfrac{1}{4}\)
※答えは帯分数に直す
\(（９）5\dfrac{1}{2}\ ×\ 1\dfrac{3}{11}\ × 2\dfrac{2}{7}\ =\)
- 答え（９）
- 16
- 解き方（９）
- \(5\dfrac{1}{2}\) × \(1\dfrac{3}{11}\) × \(2\dfrac{2}{7}\)
= \(\dfrac{\cancelto{1}{11}}{\cancelto{1}{2}}\) × \(\dfrac{\cancelto{2}{14}}{\cancelto{1}{11}}\) × \(\dfrac{\cancelto{8}{16}}{\cancelto{1}{7}}\)
※帯分数は仮分数に直す
※かけ算する前に約分する
= \(\dfrac{1\ ×\ 2\ ×\ 8}{1\ ×\ 1\ ×\ 1}\) = \(\dfrac{16}{1}\) = 16
\(（１０）6\dfrac{3}{7}\ ×\ 2\dfrac{2}{5}\ ×\ 2\dfrac{5}{8}\ =\)
- 答え（１０）
- \(40\dfrac{1}{2}\)
- 解き方（１０）
- \(6\dfrac{3}{7}\) × \(2\dfrac{2}{5}\) × \(2\dfrac{5}{8}\)
= \(\dfrac{\cancelto{9}{45}}{\cancelto{1}{7}}\) × \(\dfrac{\cancelto{3}{12}}{\cancelto{1}{5}}\) × \(\dfrac{\cancelto{3}{21}}{\cancelto{2}{8}}\)
※帯分数は仮分数に直す
※かけ算する前に約分する
= \(\dfrac{9\ ×\ 3\ ×\ 3}{1\ ×\ 1\ ×\ 2}\) = \(\dfrac{81}{2}\) = \(40\dfrac{1}{2}\)
※答えは帯分数に直す
問４ わり算
\(\dfrac{a}{b}\ ÷\ \dfrac{c}{d}\ =\ \dfrac{a}{b}\ ×\ \dfrac{d}{c}\)
- 解説
- \(\dfrac{a}{b}\) ÷ \(\dfrac{c}{d}\)
= ( \(\dfrac{a}{b}\) × \(\dfrac{d}{c}\) ) ÷ ( \(\dfrac{c}{d}\) × \(\dfrac{d}{c}\) ) わられる数、わる数に同じ数をかけても商は変わらない
= \(\dfrac{a}{b}\) × \(\dfrac{d}{c}\) ÷ 1 「÷ 1」をしても計算結果は変わらない
= \(\dfrac{a}{b}\) × \(\dfrac{d}{c}\)
\(（１）\dfrac{5}{8}\ ÷\ \dfrac{5}{16}\ =\)
- 答え（１）
- 2
- 解き方（１）
- \(\dfrac{5}{8}\) ÷ \(\dfrac{5}{16}\)
= \(\dfrac{5}{8}\) × \(\dfrac{16}{5}\)
※かけ算の形にする
= \(\dfrac{\cancelto{1}{5}}{\cancelto{1}{8}}\) × \(\dfrac{\cancelto{2}{16}}{\cancelto{1}{5}}\) = \(\dfrac{1\ ×\ 2}{1\ ×\ 1}\) = 2
\(（２）\dfrac{1}{4}\ ÷\ \dfrac{1}{6}\ =\)
- 答え（２）
- \(1\dfrac{1}{2}\)
- 解き方（２）
- \(\dfrac{1}{4}\) ÷ \(\dfrac{1}{6}\)
= \(\dfrac{1}{4}\) × \(\dfrac{6}{1}\)
※かけ算の形にする
= \(\dfrac{1}{\cancelto{2}{4}}\) × \(\dfrac{\cancelto{3}{6}}{1}\) = \(\dfrac{1\ ×\ 3}{2\ ×\ 1}\) = \(\dfrac{3}{2}\) = \(1\dfrac{1}{2}\)
\(（３）\dfrac{6}{7}\ ÷\ 3\ =\)
- 答え（３）
- \(\dfrac{2}{7}\)
- 解き方（３）
- \(\dfrac{6}{7}\) ÷ 3
= \(\dfrac{6}{7}\) ÷ \(\dfrac{3}{1}\)
= \(\dfrac{6}{7}\) × \(\dfrac{1}{3}\)
※かけ算の形にする
= \(\dfrac{\cancelto{2}{6}\ ×\ 1}{7\ ×\ \cancelto{1}{3}}\) = \(\dfrac{2}{7}\)
\(（４）5\dfrac{3}{5}\ ÷\ 2\dfrac{1}{10}\ =\)
- 答え（４）
- \(2\dfrac{2}{3}\)
- 解き方（４）
- \(5\dfrac{3}{5}\) ÷ \(2\dfrac{1}{10}\)
= \(\dfrac{28}{5}\) ÷ \(\dfrac{21}{10}\)
※帯分数は仮分数に直す
= \(\dfrac{28}{5}\) × \(\dfrac{10}{21}\)
※かけ算の形にする
= \(\dfrac{\cancelto{4}{28}\ ×\ \cancelto{2}{10}}{\cancelto{1}{5}\ ×\ \cancelto{3}{21}}\) = \(\dfrac{8}{3}\) = \(2\dfrac{2}{3}\)
\(（５）42\dfrac{2}{3}\ ÷\ 3\dfrac{7}{15}\ =\)
- 答え（５）
- \(12\dfrac{4}{13}\)
- 解き方（５）
- \(42\dfrac{2}{3}\) ÷ \(3\dfrac{7}{15}\)
= \(\dfrac{128}{3}\) ÷ \(\dfrac{52}{15}\)
※帯分数は仮分数に直す
= \(\dfrac{128}{3}\) × \(\dfrac{15}{52}\)
※かけ算の形にする
= \(\dfrac{\cancelto{32}{128}\ ×\ \cancelto{5}{15}}{\cancelto{1}{3}\ ×\ \cancelto{13}{52}}\) = \(\dfrac{160}{13}\) = \(12\dfrac{4}{13}\)
\(（６）\dfrac{1}{5}\ ÷\ \dfrac{2}{7}\ ÷\ \dfrac{4}{9}\ =\)
- 答え（６）
- \(1\dfrac{23}{40}\)
- 解き方（６）
- \(\dfrac{1}{5}\) ÷ \(\dfrac{2}{7}\) ÷ \(\dfrac{4}{9}\)
= \(\dfrac{1}{5}\) × \(\dfrac{7}{2}\) × \(\dfrac{9}{4}\) = \(\dfrac{63}{40}\) = \(1\dfrac{23}{40}\)
※かけ算の形にする
\(（７）6\dfrac{2}{3}\ ÷\ 7\dfrac{1}{5}\ ÷\ 5\dfrac{5}{6}\ =\)
- 答え（７）
- \(\dfrac{10}{63}\)
- 解き方（７）
- \(6\dfrac{2}{3}\) ÷ \(7\dfrac{1}{5}\) ÷ \(5\dfrac{5}{6}\)
= \(\dfrac{20}{3}\) ÷ \(\dfrac{36}{5}\) ÷ \(\dfrac{35}{6}\)
※帯分数は仮分数に直す
= \(\dfrac{\cancelto{10}{20}}{3}\) × \(\dfrac{\cancelto{1}{5}}{\cancelto{\cancelto{3}{6}}{36}}\) × \(\dfrac{\cancelto{1}{6}}{\cancelto{7}{35}}\) = \(\dfrac{10\ ×\ 1\ ×\ 1}{3\ ×\ 3\ × 7}\) = \(\dfrac{10}{63}\)
※かけ算の形にする
\(（８）2\dfrac{2}{13}\ ÷\ \dfrac{4}{13}\ ×\ 3\dfrac{1}{3}\ =\)
- 答え（８）
- \(23\dfrac{1}{3}\)
- 解き方（８）
- \(2\dfrac{2}{13}\) ÷ \(\dfrac{4}{13}\) × \(3\dfrac{1}{3}\)
= \(\dfrac{28}{13}\) ÷ \(\dfrac{4}{13}\) × \(\dfrac{10}{3}\)
※帯分数は仮分数に直す
= \(\dfrac{\cancelto{7}{28}}{\cancelto{1}{13}}\) × \(\dfrac{\cancelto{1}{13}}{\cancelto{1}{4}}\) × \(\dfrac{10}{3}\) = \(\dfrac{7\ ×\ 1\ ×\ 10}{1\ ×\ 1\ × 3}\) = \(\dfrac{70}{3}\) = \(23\dfrac{1}{3}\)
※わり算の部分をかけ算の形にする
\(（９）(\ 3\ ÷\ 11\ )\ ×\ 2\dfrac{4}{9}\ ÷\ \dfrac{7}{15}\ =\)
- 答え（９）
- \(1\dfrac{3}{7}\)
- 解き方（９）
- (3 ÷ 11) × \(2\dfrac{4}{9}\) ÷ \(\dfrac{7}{15}\)
= ( \(\dfrac{3}{1}\) ÷ \(\dfrac{11}{1}\) ) × \(\dfrac{22}{9}\) ÷ \(\dfrac{7}{15}\)
※帯分数は仮分数に直す
= ( \(\dfrac{3}{1}\) × \(\dfrac{1}{11}\) ) × \(\dfrac{22}{9}\) × \(\dfrac{15}{7}\)
※わり算の部分をかけ算の形にする
= \(\dfrac{\cancelto{1}{3}}{\cancelto{1}{11}}\) × \(\dfrac{\cancelto{2}{22}}{\cancelto{\cancelto{1}{3}}{9}}\) × \(\dfrac{\cancelto{5}{15}}{7}\) = \(\dfrac{1\ ×\ 2\ ×\ 5}{1\ ×\ 1\ × 7}\) = \(\dfrac{10}{7}\) = \(1\dfrac{3}{7}\)
\(（１０）2\dfrac{7}{9}\ ÷\ 3\dfrac{4}{7}\ ×\ \dfrac{2}{21}\ ÷\ \dfrac{18}{35}\ =\)
- 答え（１０）
- \(\dfrac{35}{243}\)
- 解き方（１０）
- \(2\dfrac{7}{9}\) ÷ \(3\dfrac{4}{7}\) × \(\dfrac{2}{21}\) ÷ \(\dfrac{18}{35}\)
= \(\dfrac{25}{9}\) ÷ \(\dfrac{25}{7}\) × \(\dfrac{2}{21}\) ÷ \(\dfrac{18}{35}\)
※帯分数は仮分数に直す
= \(\dfrac{\cancelto{1}{25}}{9}\) × \(\dfrac{\cancelto{1}{7}}{\cancelto{1}{25}}\) × \(\dfrac{\cancelto{1}{2}}{\cancelto{3}{21}}\) × \(\dfrac{35}{\cancelto{9}{18}}\) = \(\dfrac{1\ ×\ 1\ ×\ 1\ ×\ 35}{9\ ×\ 1\ × 3\ ×\ 9}\) = \(\dfrac{35}{243}\)
※わり算の部分をかけ算の形にする