数学Ⅰ｜数と式｜問題一覧 ＞ 問題解説 08
08｜対称式を用いた式の値
08
\(x=\displaystyle \frac{\,1\,}{\,\sqrt{3}+\sqrt{2}\,}~,~\)\(y=\displaystyle \frac{\,1\,}{\,\sqrt{3}-\sqrt{2}\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x+y=\fbox{ア}\,\sqrt{\,\fbox{イ}\,}\) \({\small (2)}~\)\(xy=\fbox{ウ}\)
\({\small (3)}~\)\(x^2y+xy^2=\fbox{エ}\,\sqrt{\,\fbox{オ}\,}\)
\({\small (4)}~\)\(x^2+y^2=\fbox{カキ}\)
\({\small (5)}~\)\(\displaystyle \frac{\,x\,}{\,y\,}+\displaystyle \frac{\,y\,}{\,x\,}=\fbox{クケ}\)
\(x=\displaystyle \frac{\,1\,}{\,\sqrt{3}+\sqrt{2}\,}~,~\)\(y=\displaystyle \frac{\,1\,}{\,\sqrt{3}-\sqrt{2}\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x+y=\fbox{ア}\,\sqrt{\,\fbox{イ}\,}\) \({\small (2)}~\)\(xy=\fbox{ウ}\)
\({\small (3)}~\)\(x^2y+xy^2=\fbox{エ}\,\sqrt{\,\fbox{オ}\,}\)
\({\small (4)}~\)\(x^2+y^2=\fbox{カキ}\)
\({\small (5)}~\)\(\displaystyle \frac{\,x\,}{\,y\,}+\displaystyle \frac{\,y\,}{\,x\,}=\fbox{クケ}\)
共通テスト数学ⅠA｜数と式
08 数学ⅠA【解答】
\({\small (1)}~\)\(x+y=2\sqrt{3}\)
\({\small (2)}~\)\(xy=1\)
\({\small (3)}~\)\(x^2y+xy^2=2\sqrt{3}\)
\({\small (4)}~\)\(x^2+y^2=10\)
\({\small (5)}~\)\(\displaystyle \frac{\,x\,}{\,y\,}+\displaystyle \frac{\,y\,}{\,x\,}=10\)
【より詳しい解説】
\(x\) と \(y\) をそれぞれ有理化すると、
\(\begin{eqnarray}~~~x&=&\displaystyle \frac{\,1\,}{\,\sqrt{3}+\sqrt{2}\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{3}-\sqrt{2}\,}{\,\sqrt{3}-\sqrt{2}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{3}-\sqrt{2}\,}{\,(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{3}-\sqrt{2}\,}{\,(\sqrt{3})^2-(\sqrt{2})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{3}-\sqrt{2}\,}{\,3-2\,}
\\[3pt]~~~&=&\sqrt{3}-\sqrt{2}\end{eqnarray}\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,1\,}{\,\sqrt{3}-\sqrt{2}\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{3}+\sqrt{2}\,}{\,\sqrt{3}+\sqrt{2}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{3}+\sqrt{2}\,}{\,(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{3}+\sqrt{2}\,}{\,(\sqrt{3})^2-(\sqrt{2})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{3}+\sqrt{2}\,}{\,3-2\,}
\\[3pt]~~~&=&\sqrt{3}+\sqrt{2}\end{eqnarray}\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~x+y&=&(\sqrt{3}-\sqrt{2})+(\sqrt{3}+\sqrt{2})
\\[3pt]~~~&=&2\sqrt{3}~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~xy&=&(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})
\\[3pt]~~~&=&(\sqrt{3})^2-(\sqrt{2})^2
\\[3pt]~~~&=&3-2
\\[3pt]~~~&=&1~~~\cdots {\small [\,2\,]}\end{eqnarray}\)
\({\small (3)}~\)
\(\begin{eqnarray}~~~x^2y+xy^2&=&xy(x+y)
\\[3pt]~~~&=&1 \cdot 2\sqrt{3}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&2\sqrt{3}\end{eqnarray}\)
\({\small (4)}~\)
\(\begin{eqnarray}~~~x^2+y^2&=&(x+y)^2-2xy
\\[3pt]~~~&=&(2\sqrt{3})^2-2 \cdot 1\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&4 \cdot 3-2
\\[3pt]~~~&=&12-2
\\[3pt]~~~&=&10\end{eqnarray}\)
\({\small (5)}~\)
\(\begin{eqnarray}~~~\displaystyle \frac{\,x\,}{\,y\,}+\displaystyle \frac{\,y\,}{\,x\,}&=&\displaystyle \frac{\,x^2+y^2\,}{\,xy\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(x+y)^2-2xy\,}{\,xy\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(2\sqrt{3})^2-2 \cdot 1\,}{\,1\,}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&4 \cdot 3-2
\\[3pt]~~~&=&12-2
\\[3pt]~~~&=&10\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,(x+y)^2-2xy\,}{\,xy\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(2\sqrt{3})^2-2 \cdot 1\,}{\,1\,}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&4 \cdot 3-2
\\[3pt]~~~&=&12-2
\\[3pt]~~~&=&10\end{eqnarray}\)
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