21102中２・式の計算・計算問題・分数をふくむ式2

$\begin{eqnarray}(1)\quad\;\;\dfrac{\;x\;}{\;3\;} + \dfrac{\;x+y\;}{\;4\;} - \dfrac{\;2x-3y\;}{\;6\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;4x\;}{\;12\;} + \dfrac{\;3(\,x+y\,)\;}{\;12\;} - \dfrac{\;2(\,2x-3y\,)\;}{\;12\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;4x+3(\,x+y\,)- 2(\,2x-3y\,)\;}{\;12\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;4x+3x+3y-4x+6y\;}{\;12\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;4x+3x-4x+3y+6y\;}{\;12\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;\color{red}\cancelto{1}{\color{black}3}\color{black}x\color{black}+\color{red}\cancelto{3}{\color{black}9}\color{black}y\;}{\;\color{red}\cancelto{4}{\color{black}12}\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;x+3y\;}{\;4\;}\end{eqnarray}\;\;$

答$\dfrac{\;x+3y\;}{\;4\;}$

$\begin{eqnarray}(2)\quad\;\;\dfrac{\;3x+6y\;}{\;5\;} + \dfrac{\;5x-3y\;}{\;4\;} - \dfrac{\;6x-3y\;}{\;10\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;4(\,3x+6y\,)\;}{\;20\;} + \dfrac{\;5(\,5x-3y\,)\;}{\;20\;} - \dfrac{\;2(\,6x-3y\,)\;}{\;20\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;4(\,3x+6y\,)+5(\,5x-3y\,)-2(\,6x-3y\,)\;}{\;20\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;12x+24y+25x-15y-12x+6y\;}{\;20\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;12x+25x-12x+24y-15y+6y\;}{\;20\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;\color{red}\cancelto{5}{\color{black}25}\color{black}x\color{black}+\color{red}\cancelto{3}{\color{black}15}\color{black}y\;}{\;\color{red}\cancelto{4}{\color{black}20}\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;5x+3y\;}{\;4\;}\end{eqnarray}\;\;$

答$\dfrac{\;5x+3y\;}{\;4\;}$

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

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;2(\,4a-3\,)\;}{\;10\;}-\dfrac{\;5(\,5b+3\,)\;}{\;10\;}-\dfrac{\;3a \times 10\;}{10}+\dfrac{\;2b \times 10\;}{10}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;2(\,4a+3\,)-5(\,5b+3\,)-30a+20b\;}{\;10\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;8a+6-25b-15-30a+20b\;}{\;10\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;8a-30a-25b+20b+6-15\;}{\;10\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;-22a-5b-9\;}{\;10\;}\end{eqnarray}\;\;$

答$\dfrac{\;-22a-5b-9\;}{\;10\;}$

$\begin{eqnarray}(4)\quad\;\;2(\,x^{2}-3x-6\,)-\dfrac{\;3x^{2}-4x+5\;}{\;2\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;2 \times 2(\,x^{2}-3x-6\,)\;}{2}-\dfrac{\;3x^{2}-4x+5\;}{\;2\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;4(\,x^{2}-3x-6\,)-(\,3x^{2}-4x+5\,)\;}{\;2\;}\end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;4x^{2}-12x-24-3x^{2}+4x-5\;}{\;2\;} \end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;4x^{2}-3x^{2}-12x+4x-24-5\;}{\;2\;} \end{eqnarray}\;\;$

$\begin{eqnarray}\quad\;\;&=&\dfrac{\;x^{2}-8x-29\;}{\;2\;}\end{eqnarray}\;\;$

答$\dfrac{\;x^{2}-8x-29\;}{\;2\;}$