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逆数との和x+1/xと対称式の値 で確認できます。
問題アーカイブ01
問題アーカイブ01\(x=\displaystyle \frac{\,3-\sqrt{5}\,}{\,2\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x+\displaystyle \frac{\,1\,}{\,x\,}\) \({\small (2)}~\)\(x^2+\displaystyle \frac{\,1\,}{\,x^2\,}\)
\({\small (3)}~\)\(x^2-\displaystyle \frac{\,1\,}{\,x^2\,}\) \({\small (4)}~\)\(x^4+\displaystyle \frac{\,1\,}{\,x^4\,}\)
\({\small (5)}~\)\(x^4-\displaystyle \frac{\,1\,}{\,x^4\,}\)
\({\small (1)}~\)\(x+\displaystyle \frac{\,1\,}{\,x\,}\) \({\small (2)}~\)\(x^2+\displaystyle \frac{\,1\,}{\,x^2\,}\)
\({\small (3)}~\)\(x^2-\displaystyle \frac{\,1\,}{\,x^2\,}\) \({\small (4)}~\)\(x^4+\displaystyle \frac{\,1\,}{\,x^4\,}\)
\({\small (5)}~\)\(x^4-\displaystyle \frac{\,1\,}{\,x^4\,}\)
数研出版｜数学Ⅰ[712] p.48 演習問題A 4
\(x=\displaystyle \frac{\,3-\sqrt{5}\,}{\,2\,}\) より、
\(\begin{eqnarray}~~~\displaystyle \frac{\,1\,}{\,x\,}&=&\displaystyle \frac{\,2\,}{\,3-\sqrt{5}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2\,}{\,3-\sqrt{5}\,}{\, \small \times \,}\displaystyle \frac{\,3+\sqrt{5}\,}{\,3+\sqrt{5}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2(3+\sqrt{5})\,}{\,(3-\sqrt{5})(3+\sqrt{5})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2(3+\sqrt{5})\,}{\,3^2-(\sqrt{5})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2(3+\sqrt{5})\,}{\,9-5\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2(3+\sqrt{5})\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,3+\sqrt{5}\,}{\,2\,}\end{eqnarray}\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~x+\displaystyle \frac{\,1\,}{\,x\,}&=&\displaystyle \frac{\,3-\sqrt{5}\,}{\,2\,}+\displaystyle \frac{\,3+\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(3-\sqrt{5})+(3+\sqrt{5})\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,6\,}{\,2\,}
\\[5pt]~~~&=&3~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\(\begin{eqnarray}~~~x-\displaystyle \frac{\,1\,}{\,x\,}&=&\displaystyle \frac{\,3-\sqrt{5}\,}{\,2\,}-\displaystyle \frac{\,3+\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(3-\sqrt{5})-(3+\sqrt{5})\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,-2\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&-\sqrt{5}~~~\cdots {\small [\,2\,]}\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~&&x^2+\displaystyle \frac{\,1\,}{\,x^2\,}
\\[5pt]~~~&=&\left(x+\displaystyle \frac{\,1\,}{\,x\,}\right)^2-2 \cdot x \cdot \displaystyle \frac{\,1\,}{\,x\,}
\\[5pt]~~~&=&3^2-2\hspace{30pt}(\,∵~{\small [\,1\,]}\,)
\\[3pt]~~~&=&9-2
\\[3pt]~~~&=&7~~~\cdots {\small [\,3\,]}\end{eqnarray}\)
\({\small (3)}~\)
\(\begin{eqnarray}~~~&&x^2-\displaystyle \frac{\,1\,}{\,x^2\,}
\\[5pt]~~~&=&\left(x+\displaystyle \frac{\,1\,}{\,x\,}\right)\left(x-\displaystyle \frac{\,1\,}{\,x\,}\right)
\\[5pt]~~~&=&3{\, \small \times \,}(-\sqrt{5})\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&-3\sqrt{5}~~~\cdots {\small [\,4\,]}\end{eqnarray}\)
\({\small (4)}~\)
\(\begin{eqnarray}~~~&&x^4+\displaystyle \frac{\,1\,}{\,x^4\,}
\\[5pt]~~~&=&\left(x^2+\displaystyle \frac{\,1\,}{\,x^2\,}\right)^2-2 \cdot x^2 \cdot \displaystyle \frac{\,1\,}{\,x^2\,}
\\[5pt]~~~&=&7^2-2\hspace{30pt}(\,∵~{\small [\,3\,]}\,)
\\[3pt]~~~&=&49-2
\\[3pt]~~~&=&47\end{eqnarray}\)
\({\small (5)}~\)
\(\begin{eqnarray}~~~&&x^4-\displaystyle \frac{\,1\,}{\,x^4\,}
\\[5pt]~~~&=&\left(x^2+\displaystyle \frac{\,1\,}{\,x^2\,}\right)\left(x^2-\displaystyle \frac{\,1\,}{\,x^2\,}\right)
\\[5pt]~~~&=&7{\, \small \times \,}(-3\sqrt{5})\hspace{30pt}(\,∵~{\small [\,3\,]}~,~{\small [\,4\,]}\,)
\\[3pt]~~~&=&-21\sqrt{5}\end{eqnarray}\)
問題アーカイブ02
問題アーカイブ02\(x=\displaystyle \frac{\,\sqrt{5}-\sqrt{3}\,}{\,\sqrt{5}+\sqrt{3}\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x+\displaystyle \frac{\,1\,}{\,x\,}\) \({\small (2)}~\)\(x^2+\displaystyle \frac{\,1\,}{\,x^2\,}\)
\({\small (1)}~\)\(x+\displaystyle \frac{\,1\,}{\,x\,}\) \({\small (2)}~\)\(x^2+\displaystyle \frac{\,1\,}{\,x^2\,}\)
東京書籍｜Advanced数学Ⅰ[701] p.48 練習問題A 5
東京書籍｜Standard数学Ⅰ[702] p.51 Level Up 6
\(x=\displaystyle \frac{\,\sqrt{5}-\sqrt{3}\,}{\,\sqrt{5}+\sqrt{3}\,}\) より、
\(\displaystyle \frac{\,1\,}{\,x\,}=\displaystyle \frac{\,\sqrt{5}+\sqrt{3}\,}{\,\sqrt{5}-\sqrt{3}\,}\)
\(x\) と \(\displaystyle \frac{\,1\,}{\,x\,}\) をそれぞれ有理化すると、
\(\begin{eqnarray}~~~x&=&\displaystyle \frac{\,\sqrt{5}-\sqrt{3}\,}{\,\sqrt{5}+\sqrt{3}\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{5}-\sqrt{3}\,}{\,\sqrt{5}-\sqrt{3}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{5}-\sqrt{3})^2\,}{\,(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{5})^2-2\sqrt{5}\cdot\sqrt{3}+(\sqrt{3})^2\,}{\,(\sqrt{5})^2-(\sqrt{3})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,5-2\sqrt{15}+3\,}{\,5-3\,}
\\[5pt]~~~&=&\displaystyle \frac{\,8-2\sqrt{15}\,}{\,2\,}
\\[5pt]~~~&=&4-\sqrt{15}\end{eqnarray}\)
\(\begin{eqnarray}~~~\displaystyle \frac{\,1\,}{\,x\,}&=&\displaystyle \frac{\,\sqrt{5}+\sqrt{3}\,}{\,\sqrt{5}-\sqrt{3}\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{5}+\sqrt{3}\,}{\,\sqrt{5}+\sqrt{3}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{5}+\sqrt{3})^2\,}{\,(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{5})^2+2\sqrt{5}\cdot\sqrt{3}+(\sqrt{3})^2\,}{\,(\sqrt{5})^2-(\sqrt{3})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,5+2\sqrt{15}+3\,}{\,5-3\,}
\\[5pt]~~~&=&\displaystyle \frac{\,8+2\sqrt{15}\,}{\,2\,}
\\[5pt]~~~&=&4+\sqrt{15}\end{eqnarray}\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~x+\displaystyle \frac{\,1\,}{\,x\,}&=&(4-\sqrt{15})+(4+\sqrt{15})
\\[5pt]~~~&=&8~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~&&x^2+\displaystyle \frac{\,1\,}{\,x^2\,}
\\[5pt]~~~&=&\left(x+\displaystyle \frac{\,1\,}{\,x\,}\right)^2-2 \cdot x \cdot \displaystyle \frac{\,1\,}{\,x\,}
\\[5pt]~~~&=&8^2-2\hspace{30pt}(\,∵~{\small [\,1\,]}\,)
\\[3pt]~~~&=&64-2
\\[3pt]~~~&=&62\end{eqnarray}\)
