21032中２・式の計算・計算問題・多項式の減法2

$\begin{eqnarray}(1)\quad\;\;(\,1.9x-0.4y-0.8\,)-(\,-0.8x-0.4y-1.9\,)\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&1.9x-0.4y-0.8\color{red}+\color{black}0.8x\color{red}+\color{black}0.4y\color{red}+\color{black}1.9\\[5pt]&=&1.9x+0.8x-0.4y+0.4y-0.8+1.9\\[5pt]&=&2.7x+1.1\end{eqnarray}\;\;$

答$2.7x+1.1$

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

$\begin{eqnarray}\quad\;\;&=&-\dfrac{\;a^{2}\;}{\;4\;}+\dfrac{\;a\;}{\;3\;}-\dfrac{\;1\;}{\;2\;}\color{red}-\color{black}\dfrac{\;a^{2}\;}{\;2\;}\color{red}-\color{black}\dfrac{a}{\;3\;}\color{red}-\color{black}\dfrac{\;3\;}{\;4\;}\\[6pt]&=&-\dfrac{\;a^{2}\;}{\;4\;}-\dfrac{\;a^{2}\;}{\;2\;}+\dfrac{\;a\;}{\;3\;}-\dfrac{a}{\;3\;}-\dfrac{\;1\;}{\;2\;}-\dfrac{\;3\;}{\;4\;}\\[6pt]&=&-\dfrac{\;1\;}{\;4\;}a^{2}-\dfrac{\;2\;}{\;4\;}a^{2}-\dfrac{\;2\;}{\;4\;}-\dfrac{\;3\;}{\;4\;}\\[6pt]&=&-\dfrac{\;3\;}{\;4\;}a^{2}-\dfrac{\;5\;}{\;4\;}\end{eqnarray}\;\;$

答$-\dfrac{\;3\;}{\;4\;}a^{2}-\dfrac{\;5\;}{\;4\;}$

$\begin{eqnarray}(3)\quad\;\;(\,0.16a^{2}+0.1ab-1.7b^{2}\,)-(\,-0.04a^{2}+0.02ab-0.3b^{2}\,)\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&0.16a^{2}+0.1ab-1.7b^{2}\color{red}+\color{black}0.04a^{2}\color{red}-\color{black}0.02ab\color{red}+\color{black}0.3b^{2}\\[5pt]&=&0.16a^{2}+0.04a^{2}+0.1ab-0.02ab-1.7b^{2}+0.3b^{2}\\[5pt]&=&0.2a^{2}+0.08ab-1.4b^{2}\end{eqnarray}\;\;$

答$0.2a^{2}+0.08ab-1.4b^{2}$

$\begin{eqnarray}(4)\quad\;\;\Bigl(\,-x^{2}+\dfrac{\;xy\;}{\;7\;}-0.2y^{2}\,\Bigr)-\Bigl(\,-\dfrac{3}{\;4\;}x^{2}-2xy-\dfrac{\;y^{2}\;}{10}\,\Bigr)\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&-x^{2}+\dfrac{\;xy\;}{\;7\;}-0.2y^{2}\color{red}+\color{black}\dfrac{3}{\;4\;}x^{2}\color{red}+\color{black}2xy\color{red}+\color{black}\dfrac{\;y^{2}\;}{10}\\[6pt]&=&-x^{2}+\dfrac{3}{\;4\;}x^{2}+\dfrac{\;xy\;}{\;7\;}+2xy-0.2y^{2}+\dfrac{\;y^{2}\;}{10}\\[6pt]&=&-\dfrac{4}{\;4\;}x^{2}+\dfrac{3}{\;4\;}x^{2}+\dfrac{\;1\;}{\;7\;}xy+\dfrac{\;14\;}{\;7\;}xy-\dfrac{2}{\;10\;}y^{2}+\dfrac{\;1\;}{10}y^{2}\\[6pt]&=&-\dfrac{1}{\;4\;}x^{2}+\dfrac{\;15\;}{\;7\;}xy-\dfrac{1}{\;10\;}y^{2}\end{eqnarray}\;\;$

答$-\dfrac{1}{\;4\;}x^{2}+\dfrac{\;15\;}{\;7\;}xy-\dfrac{1}{\;10\;}y^{2}$