【問題】 $\textcolor{green}{x=\sqrt{3}+\sqrt{2}}$ , $\textcolor{green}{y=\sqrt{3}-\sqrt{2}}$ のとき、次の式の値を求めなさい。

代入のポイント：先に式を変形(簡単)にする

(1) $\textcolor{green}{xy}$　$\textcolor{blue}{←変形できないので、そのまま代入}$

　$=(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})$

　$=(\sqrt{3})^2-(\sqrt{2})^2=3-2=\textcolor{red}{1}$

(2) $\textcolor{green}{x^2-y^2}$　$\textcolor{blue}{←因数分解できる}$

　$=(x+y)(x-y)$

　$=2\sqrt{3}×2\sqrt{2}=\textcolor{red}{4\sqrt{6}}$

\begin{eqnarray} \textcolor{green}{(3)}& &\textcolor{green}{x^2-2xy+y^2}　\textcolor{blue}{←先に式を変形する}\\\\ &=&(x-y)^2\\\\ &=&\{(\sqrt3+\sqrt{2})-(\sqrt3-\sqrt{2})\}^2\\\\ &=&(\sqrt3+\sqrt{2}-\sqrt3-\sqrt{2})^2\\\\ &=&(2\sqrt{2})^2\\\\ &=&\textcolor{red}{8} \end{eqnarray}

【問題】$\textcolor{green}{\sqrt{3}=1.732}$ , $\textcolor{green}{\sqrt{30}=5.477}$ として、次の値を求めなさい。

\begin{eqnarray} (1)& &\sqrt{300}=10\textcolor{blue}{\sqrt{3}}=10×1.732=\textcolor{red}{17.32}\\\\ (2)& &\sqrt{120}=2\textcolor{blue}{\sqrt{30}}=2×5.477=\textcolor{red}{10.954}\\\\& &小数は分数になおす\\ (3)& &\sqrt{0.03}=\frac{\textcolor{blue}{\sqrt{3}}}{10}=1.732÷10=\textcolor{red}{0.1732}\\\\ (4)& &\sqrt{0.3}=\frac{\textcolor{blue}{\sqrt{30}}}{10}=5.477÷10=\textcolor{red}{0.5477} \end{eqnarray}