12015中１・正の数と負の数・計算問題・加法5
banno計算問題 》加法⑤
次の式を，工夫して計算しなさい。
(1) $\;\;(\,-17\,)+(\,+16\,)+(\,-3\,)$
(2) $\;\;(\,-2.4\,)+(\,+4.7\,)+(\,-2.3\,)$
(3) $\;\;(\,+47\,)+(\,-24\,)+(\,-16\,)$
(4) $\;\;\Bigl(\,-\dfrac{2}{\;5\;}\,\Bigr)+\Bigl(\,+\dfrac{1}{\;3\;}\,\Bigr)+\Bigl(\,+\dfrac{2}{\;3\;}\,\Bigr)$
(5) $\;\;(\,-5.4\,)+(\,+2.8\,)+(\,-4.6\,)+(\,-2.8\,)$
(6) $\;\;\Bigl(\,+\dfrac{1}{\;4\;}\,\Bigr)+\Bigl(\,-\dfrac{2}{\;7\;}\,\Bigr)+\Bigl(\,-\dfrac{5}{\;4\;}\,\Bigr)+\Bigl(\,+\dfrac{9}{\;7\;}\,\Bigr)$
解答・解説
$\;\;\begin{eqnarray}(1)\quad\;\;(\,-17\,)+(\,+16\,)+(\,-3\,)\end{eqnarray}\;\;$
$\qquad\begin{eqnarray}&=&(\,-17\,)+\color{red}(\,-3\,)\color{black}+\color{red}(\,+16\,)\color{black}\\[6pt]&=&\{(\,-17\,)+(\,-3\,)\}+(\,+16\,)\\[6pt]&=&(\,-20\,)+(\,+16\,)\\[6pt]&=&-\,4\end{eqnarray}\;\;$
答$-\,4$
$\;\;\begin{eqnarray}(2)\quad\;\;(\,-2.4\,)+(\,+4.7\,)+(\,-2.3\,)\end{eqnarray}\;\;$
$\qquad\begin{eqnarray}&=&(\,-2.4\,)+\color{red}(\,-2.3\,)\color{black}+\color{red}(\,+4.7\,)\color{black}\\[6pt]&=&\{(\,-2.4\,)+(\,-2.3\,)\}+(\,+4.7\,)\\[6pt]&=&(\,-4.7\,)+(\,+4.7\,)\\[6pt]&=&0\end{eqnarray}\;\;$
答$0$
$\;\;\begin{eqnarray}(3)\quad\;\;(\,+47\,)+\color{red}\{\color{black}(\,-24\,)+(\,-16\,)\color{red}\}\color{black}\end{eqnarray}\;\;$
$\qquad\begin{eqnarray}&=&(\,+47\,)+(\,-40\,)\\[6pt]&=&+\,7\end{eqnarray}\;\;$
答$+\,7$
$\;\;\begin{eqnarray}(4)\quad\;\;\Bigl(\,-\dfrac{2}{\;5\;}\,\Bigr)+\color{red}\Bigl\{\color{black}\Bigl(\,+\dfrac{1}{\;3\;}\,\Bigr)+\Bigl(\,+\dfrac{2}{\;3\;}\,\Bigr)\color{red}\Bigr\}\end{eqnarray}\;\;$
$\qquad\begin{eqnarray}&=&\Bigl(\,-\dfrac{2}{\;5\;}\,\Bigr)+\Bigl(\,+\dfrac{3}{\;3\;}\,\Bigr)\\[6pt]&=&\Bigl(\,-\dfrac{2}{\;5\;}\,\Bigr)+(\,+1\,)\\[6pt]&=&\Bigl(\,-\dfrac{2}{\;5\;}\,\Bigr)+\Bigl(\,+\dfrac{5}{\;5\;}\,\Bigr)\\[6pt]&=&+\dfrac{3}{\;5\;}\end{eqnarray}\;\;$
答$+\dfrac{3}{\;5\;}$
$\;\;\begin{eqnarray}(5)\quad\;\;(\,-5.4\,)+(\,+2.8\,)+(\,-4.6\,)+(\,-2.8\,)\end{eqnarray}\;\;$
$\qquad\begin{eqnarray}&=&(\,-5.4\,)+\color{red}(\,-4.6\,)\color{black}+\color{red}(\,+2.8\,)\color{black}+(\,-2.8\,)\\[6pt]&=&\color{red}\{\color{black}(\,-5.4\,)+(\,-4.6\,)\color{red}\}\color{black}+\color{red}\{\color{black}(\,+2.8\,)+(\,-2.8\,)\color{red}\}\color{black}\\[6pt]&=&(\,-10\,)+0\\[6pt]&=&-\,10\end{eqnarray}\;\;$
答$-\,10$
$\;\;\begin{eqnarray}(6)\quad\;\;\Bigl(\,+\dfrac{1}{\;4\;}\,\Bigr)+\Bigl(\,-\dfrac{2}{\;7\;}\,\Bigr)+\Bigl(\,-\dfrac{5}{\;4\;}\,\Bigr)+\Bigl(\,+\dfrac{9}{\;7\;}\,\Bigr)\end{eqnarray}\;\;$
$\qquad\begin{eqnarray}&=&\Bigl(\,+\dfrac{1}{\;4\;}\,\Bigr)+\color{red}\Bigl(\,-\dfrac{5}{\;4\;}\,\Bigr)\color{black}+\color{red}\Bigl(\,-\dfrac{2}{\;7\;}\,\Bigr)\color{black}+\Bigl(\,+\dfrac{9}{\;7\;}\,\Bigr)\\[6pt]&=&\color{red}\Bigl\{\color{black}\Bigl(\,+\dfrac{1}{\;4\;}\,\Bigr)+\Bigl(\,-\dfrac{5}{\;4\;}\,\Bigr)\color{red}\Bigr\}\color{black}+\color{red}\Bigl\{\color{black}\Bigl(\,-\dfrac{2}{\;7\;}\,\Bigr)+\Bigl(\,+\dfrac{9}{\;7\;}\,\Bigr)\color{red}\Bigr\}\color{black}\\[6pt]&=&\Bigl(\,-\dfrac{4}{\;4\;}\,\Bigr)+\Bigl(\,+\dfrac{7}{\;7\;}\,\Bigr)\\[6pt]&=&(\,-1\,)+(\,+1\,)\\[4pt]&=&0\end{eqnarray}\;\;$
答$0$
