31031中3・多項式の展開・計算問題・展開の公式2-1
計算問題 》展開の公式2①
次の式を展開しなさい。
(1) $(\,x+2\,)^{2}$
(2) $(\,x+10\,)^{2}$
(3) $(\,x+0.6\,)^{2}$
(4) $(\,x-1\,)^{2}$
(5) $\Bigl(x-\dfrac{2}{\;3\;}\Bigr)^{2}$
(6) $(\,4-x\,)^{2}$
解答・解説
(1) $\quad\;\,\begin{eqnarray}(\,x+2\,)^{2}\end{eqnarray}$
$\;\;\begin{eqnarray}&=&x^{2}+2\times 2\times x+2^{2}\\[6pt]&=&x^{2}+4x+4\end{eqnarray}\;\;$
答$x^{2}+4x+4$
(2) $\quad\;\,\begin{eqnarray}(\,x+10\,)^{2}\end{eqnarray}$
$\;\;\begin{eqnarray}&=&x^{2}+2\times 10\times x+10^{2}\\[6pt]&=&x^{2}+20x+100\end{eqnarray}\;\;$
答$x^{2}+20x+100$
(3) $\quad\;\,\begin{eqnarray}(\,x+0.6\,)^{2}\end{eqnarray}$
$\;\;\begin{eqnarray}&=&x^{2}+2\times 0.6\times x+0.6^{2}\\[6pt]&=&x^{2}+1.2x+0.36\end{eqnarray}\;\;$
答$x^{2}+1.2x+0.36$
(4) $\quad\;\,\begin{eqnarray}(\,x-1\,)^{2}\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}-2\times 1\times x+1^{2}\\[6pt]&=&x^{2}-2x+1\end{eqnarray}\;\;$
答$x^{2}-2x+1$
(5) $\quad\;\,\begin{eqnarray}\Bigl(\,x-\frac{2}{3}\,\Bigr)^{2}\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&x^{2}-2\times \frac{2}{\;3\;}\times x +\Bigl(\,-\frac{2}{\;3\;}\,\Bigr)^{2}\\[6pt]&=&x^{2}-\frac{4}{\;3\;}x+\frac{4}{\;9\;}\end{eqnarray}\;\;$
答$x^{2}-\dfrac{4}{\;3\;}x+\dfrac{4}{\;9\;}$
(6) $\quad\;\,\begin{eqnarray}(\,4-x\,)^{2}\end{eqnarray}\;\;$
$\;\;\begin{eqnarray}&=&4^{2}-2\times x\times 4+x^{2}\\[6pt]&=&16-8x+x^{2}\\[6pt]&=&x^{2}-8x+16\end{eqnarray}\;\;$
答$x^{2}-8x+16$