12062中１・正の数と負の数・計算問題・乗法と除法2

$\;\;\begin{eqnarray}(1)\quad\;\;(\;-3\;)^{2} \div (\,-2^{3}\,) \times (\,-8\,)\end{eqnarray}\;\;$

$\qquad\begin{eqnarray}&=&9 \div (\,-8\,) \times (\,-8\,)\\[6pt]&=&9 \times \Bigl(\,-\dfrac{1}{\;8\;}\,\Bigr) \times (\,-8\,)\\[6pt]&=&9 \times \Bigl\{\Bigl(\,-\dfrac{1}{\;8\;}\,\Bigr) \times (\,-8\,)\Bigr\}\\[6pt]&=&9 \times 1\\[6pt]&=&9\end{eqnarray}\;\;$

答$9$

$\;\;\begin{eqnarray}(2)\quad\;\;(\,-2^{2}\,) \times (\,-2\,)^{2} \div (\,-4^{2}\,)\end{eqnarray}\;\;$

$\qquad\begin{eqnarray}&=&(\,-4\,) \times 4 \div (\,-16\,)\\[6pt]&=&(\,-16\,) \div (\,-16\,)\\[6pt]&=&1\end{eqnarray}\;\;$

答$1$

$\;\;\begin{eqnarray}(3)\quad\;\;18 \div \Bigl(\,-\dfrac{3}{\;2\;}\,\Bigr)^{2} \times \Bigl(\,-\dfrac{1}{\;2\;}\,\Bigr)^{3}\end{eqnarray}\;\;$

$\qquad\begin{eqnarray}&=&18 \div \dfrac{9}{\;4\;} \times \Bigl(\,-\dfrac{1}{\;8\;}\,\Bigr)\\[6pt]&=&\color{red}\cancelto{2}{\color{black}18} \color{black}\times \dfrac{4}{\;\color{red}\cancelto{1}{\color{black}9}\;} \color{black}\times \Bigl(\,-\dfrac{1}{\;8\;}\,\Bigr)\\[6pt]&=&8\times \Bigl(\,-\dfrac{1}{\;8\;}\,\Bigr)\\[6pt]&=&-1\end{eqnarray}\;\;$

答$-\,1$

$\;\;\begin{eqnarray}(4)\quad\;\;(\,-4\,)^{2} \div \{-(\,-2\,)^{3}\} \times (\,-1^{2}\,)\end{eqnarray}\;\;$

$\qquad\begin{eqnarray}&=&16 \div \{-(\,-8\,)\} \times (\,-1\,)\\[6pt]&=&16 \div 8 \times (\,-1\,)\\[6pt]&=&2 \times (\,-1\,)\\[6pt]&=&-2\end{eqnarray}\;\;$

答$-\,2$

$\;\;\begin{eqnarray}(5)\quad\;\;\dfrac{9}{\;10\;} \times \Bigl(\,-\dfrac{1}{\;3\;} \,\Bigr)^{2} \div \Bigl\{-\Bigl(\,-\dfrac{1}{\;5\;} \,\Bigr)^{2}\Bigr\}\end{eqnarray}\;\;$

$\qquad\begin{eqnarray}&=&\dfrac{9}{\;10\;} \times \dfrac{1}{\;9\;} \div \Bigl(\,-\dfrac{1}{\;25\;} \,\Bigr)\\[6pt]&=&\dfrac{9}{\;10\;} \times \dfrac{1}{\;9\;} \times (\,-25\,)\\[6pt]&=&\dfrac{1}{\;10\;} \times (\,-25\,)\\[6pt]&=&-\dfrac{5}{\;2\;}\end{eqnarray}\;\;$

答$-\,\dfrac{5}{\;2\;}$