12033中１・正の数と負の数・計算問題・加法と減法3

$\;\;\begin{eqnarray}(1)\quad\;\;-0.4+(\,-2\,)-(\,+1.6\,)-(\,-3\,)\end{eqnarray}\;\;$

$\qquad\begin{eqnarray}&=&-0.4+(\,-2\,)+(\,-1.6\,)+(\,+3\,)\\[5pt]&=&\,\color{red}-0.4-1.6\,\color{blue}-2+3\\[5pt]&=&-2+1\\[5pt]&=&-1\end{eqnarray}\;\;$

答$-\,1$

$\;\;\begin{eqnarray}(2)\quad\;\;\frac{3}{\;4\;}-\Bigl(\,+\dfrac{1}{\;3\;}\,\Bigr)+\dfrac{1}{\;6\;}+\Bigl(\,-\dfrac{1}{\;2\;}\,\Bigr)\end{eqnarray}\;\;$

$\qquad\begin{eqnarray}&=&\frac{3}{\;4\;}+\Bigl(\,-\dfrac{1}{\;3\;}\,\Bigr)+\dfrac{1}{\;6\;}+\Bigl(\,-\dfrac{1}{\;2\;}\,\Bigr)\\[6pt]&=&\frac{3}{\;4\;}-\dfrac{1}{\;3\;}+\dfrac{1}{\;6\;}-\dfrac{1}{\;2\;}\\[6pt]&=&\frac{3}{\;4\;}-\dfrac{1}{\;3\;}+\dfrac{1}{\;6\;}-\dfrac{1}{\;2\;}\\[6pt]&=&\color{red}\frac{9}{\;12\;}-\dfrac{4}{\;12\;}\color{blue}+\dfrac{2}{\;12\;}-\dfrac{6}{\;12\;}\\[6pt]&=&\dfrac{5}{\;12\;}-\dfrac{4}{\;12\;}\\[6pt]&=&\dfrac{1}{\;12\;}\end{eqnarray}\;\;$

答$\dfrac{1}{\;12\;}$

$\;\;\begin{eqnarray}(3)\quad\;\;-\dfrac{2}{\;5\;}-\Bigl(\,+\dfrac{1}{\;4\;}\,\Bigr)-(\,-0.4\,)+1.25\end{eqnarray}\;\;$

$\qquad\begin{eqnarray}&=&-\dfrac{2}{\;5\;}+\Bigl(\,-\dfrac{1}{\;4\;}\,\Bigr)+(\,+0.4\,)+1.25\\[6pt]&=&-\dfrac{2}{\;5\;}-\dfrac{1}{\;4\;}+0.4+1.25\\[6pt]&=&-\dfrac{2}{\;5\;}-\dfrac{1}{\;4\;}+\dfrac{2}{\;5\;}+\dfrac{5}{\;4\;}\\[6pt]&=&\,\color{red}-\dfrac{2}{\;5\;}+\dfrac{2}{\;5\;}\color{blue}-\dfrac{1}{\;4\;}+\dfrac{5}{\;4\;}\\[6pt]&=&\color{red}0\color{blue}+1\\[6pt]&=&1\end{eqnarray}\;\;$

答$1$

$\;\;\begin{eqnarray}(4)\quad\;\;-\dfrac{1}{\;4\;}+0.125+0.25-\dfrac{1}{\;8\;}\end{eqnarray}\;\;$

$\qquad\begin{eqnarray}&=&-\dfrac{1}{\;4\;}+\dfrac{1}{\;8\;}+\dfrac{1}{\;4\;}-\dfrac{1}{\;8\;}\\[6pt]&=&\color{red}-\dfrac{1}{\;4\;}+\dfrac{1}{\;4\;}\color{blue}+\dfrac{1}{\;8\;}-\dfrac{1}{\;8\;}\\[5pt]&=&\color{red}0\color{blue}+0\\[5pt]&=&0\end{eqnarray}\;\;$

答$0$

$\;\;\begin{eqnarray}(5)\quad\;\;-\dfrac{5}{\;2\;}+4-0.8-\dfrac{3}{\;5\;}\end{eqnarray}\;\;$

$\qquad\begin{eqnarray}&=&\,\color{red}-\dfrac{25}{\;10\;}+\dfrac{40}{\;10\;}\color{blue}-\dfrac{8}{\;10\;}-\dfrac{6}{\;10\;}\\[5pt]&=&\color{red}\dfrac{15}{\;10\;}\color{blue}-\dfrac{14}{\;10\;}\\[6pt]&=&\dfrac{1}{\;10\;}\end{eqnarray}\;\;$

答$\dfrac{1}{\;10\;}$