31032中３・多項式の展開・計算問題・展開の公式2-2

(1) $\quad\;\,\begin{eqnarray}(\,2x+2\,)^{2}\end{eqnarray}$

$\;\;\begin{eqnarray}&=&(\,2x\,)^{2}+2\times 2\times 2x+2^{2}\\[6pt]&=&4x^{2}+8x+4\end{eqnarray}\;\;$

答$4x^{2}+8x+4$

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

$\;\;\begin{eqnarray}&=&(\,5x\,)^{2}+2\times 4 \times 5x + 4^{2}\\[6pt]&=&25x^{2}+40x+16\end{eqnarray}\;\;$

答$25x^{2}+40x+16$

(3) $\quad\;\,\begin{eqnarray}(\,0.2x+5\,)^{2}\end{eqnarray}$

$\;\;\begin{eqnarray}&=&(\,0.2x\,)^{2}+2\times 5 \times 0.2x + 5^{2}\\[6pt]&=&0.04x^{2}+2x+25\end{eqnarray}\;\;$

答$0.04x^{2}+2x+25$

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

$\;\;\begin{eqnarray}&=&(\,4x\,)^{2}-2\times 1 \times 4x+1^{2}\\[6pt]&=&16x^{2}-8x+1\end{eqnarray}\;\;$

答$16x^{2}-8x+1$

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

$\;\;\begin{eqnarray}&=&\Bigl(\dfrac{\;1\;}{\;2\;}x\Bigr)^{2}-2\times \dfrac{\;1\;}{\;3\;}\times \dfrac{\;1\;}{\;2\;}x +\Bigl(\,\dfrac{1}{\;3\;}\,\Bigr)^{2}\\[6pt]&=&\dfrac{\;1\;}{\;4\;}x^{2}- \dfrac{\;1\;}{\;3\;}x +\dfrac{1}{\;9\;}\end{eqnarray}\;\;$

答$\dfrac{\;1\;}{\;4\;}x^{2}- \dfrac{\;1\;}{\;3\;}x +\dfrac{1}{\;9\;}$

(6) $\quad\;\,\begin{eqnarray}(\,10-0.1x\,)^{2}\end{eqnarray}\;\;$

$\;\;\begin{eqnarray}&=&10^{2}-2\times 0.1x \times 10+(\,0.1x\,)^{2}\\[6pt]&=&100-2x+0.01x^{2}\\[6pt]&=&0.01x^{2}-2x+100\end{eqnarray}\;\;$

答$0.01x^{2}-2x+100$