問題
次の計算をしなさい. 解答は答えのみでよい.
解説
(1) $\,1\div\{225\div(2024\times2025)\times253\}$
$1\div\bigg(\dfrac{225\times253}{2024\times2025}\bigg)$
$=1\div\bigg(\dfrac{1\times1}{8\times9}\bigg)$
$=$$72$
(2) $\bigg\{\bigg(3\dfrac{1}{2}+\dfrac{2}{5}\bigg)\div4\dfrac{1}{3}-\dfrac{2}{3}\times1\dfrac{1}{8}\bigg\}\div2.25$
$\bigg(\dfrac{39}{10}\times\dfrac{3}{13}-\dfrac{2}{3}\times\dfrac{9}{8}\bigg)\times\dfrac{4}{9}$
$=\bigg(\dfrac{9}{10}-\dfrac{3}{4}\bigg)\times\dfrac{4}{9}$
$=\dfrac{3}{20}\times\dfrac{4}{9}$
$=$$\dfrac{1}{15}$
(3) $\,3.14\times5.7+3.14\times4.8-0.314\times5$
$3.14\times(5.7+4.8-0.5)$
$=3.14\times10$
$=$$31.4$
(4) $\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}$
通分して計算しても良いが, $\dfrac{1}{n\times(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$を利用すると,
$\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}$
$=1-\dfrac{1}{7}$
$=$$\dfrac{6}{7}$
と計算できます.
