問題
次の◻︎に当てはまる数を求めなさい.
- (1) $20-(17-7)\times2=◻︎$
- $20-(17-7)\times2=20-10\times2=0$より, 0になります.
- (2) $0.24\div0.015\times5=◻︎$
- $0.24\div0.015\times5=\dfrac{24}{100}\div\dfrac{15}{1000}\times5=\dfrac{24}{100}\times\dfrac{1000}{15}\times5=80$より, 80になります.
- (3) $3\dfrac{1}{2}\times\bigg(\dfrac{1}{2}-\dfrac{1}{7}\bigg)\div1\dfrac{1}{4}=◻︎$
- $3\dfrac{1}{2}\times\bigg(\dfrac{1}{2}-\dfrac{1}{7}\bigg)\div1\dfrac{1}{4}=\dfrac{7}{2}\times\dfrac{5}{14}\times\dfrac{4}{5}=1$より, 1になります.
- (4) $15-2.4\times\dfrac{3}{8}\div5\times\dfrac{5}{6}=◻︎$
- $15-2.4\times\dfrac{3}{8}\div5\times\dfrac{5}{6}=15-\dfrac{24}{10}\times\dfrac{3}{8}\times\dfrac{1}{5}\times\dfrac{5}{6}=14\dfrac{17}{20}$より, $14\dfrac{17}{20}$となります.
- (5) $◻︎\times0.5-0.3\times\dfrac{5}{6}=\dfrac{7}{12}$
- $◻︎\times\dfrac{1}{2}-\dfrac{3}{10}\times\dfrac{5}{6}=\dfrac{7}{12}$
$◻︎\times\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{7}{12}$
$◻︎\times\dfrac{1}{2}=\dfrac{7}{12}+\dfrac{1}{4}$
$◻︎\times\dfrac{1}{2}=\dfrac{5}{6}$
$◻︎=\dfrac{5}{6}\times2=1\dfrac{2}{3}$より, $1\dfrac{2}{3}$になります.
- $◻︎\times\dfrac{1}{2}-\dfrac{3}{10}\times\dfrac{5}{6}=\dfrac{7}{12}$
- (6) $1\dfrac{7}{23}\times\dfrac{1}{3}+\dfrac{1}{4}\times1\dfrac{7}{23}-1\dfrac{7}{23}\times\dfrac{1}{5}=◻︎$
- $1\dfrac{7}{23}\times\dfrac{1}{3}+\dfrac{1}{4}\times1\dfrac{7}{23}-1\dfrac{7}{23}\times\dfrac{1}{5}=\dfrac{30}{23}\times\bigg(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}\bigg)=\dfrac{30}{23}\times\dfrac{23}{60}=\dfrac{1}{2}$より, $\dfrac{1}{2}$になります.
