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交代式や3次式の対称式の値 で確認できます。
問題アーカイブ01
問題アーカイブ01\(x=\displaystyle \frac{\,1\,}{\,\sqrt{2}+1\,}~,~\)\(y=\displaystyle \frac{\,1\,}{\,\sqrt{2}-1\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x^2+y^2\)
\({\small (2)}~\)\(x^3+y^3\)
\({\small (3)}~\)\(x^3y-x^2y^2+xy^3\)
\({\small (1)}~\)\(x^2+y^2\)
\({\small (2)}~\)\(x^3+y^3\)
\({\small (3)}~\)\(x^3y-x^2y^2+xy^3\)
数研出版｜数学Ⅰ[712] p.35 発展 練習1
\(x\) と \(y\) をそれぞれ有理化すると、
\(\begin{eqnarray}~~~x&=&\displaystyle \frac{\,1\,}{\,\sqrt{2}+1\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{2}-1\,}{\,\sqrt{2}-1\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{2}-1\,}{\,(\sqrt{2}+1)(\sqrt{2}-1)\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{2}-1\,}{\,(\sqrt{2})^2-1^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{2}-1\,}{\,2-1\,}
\\[3pt]~~~&=&\sqrt{2}-1\end{eqnarray}\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,1\,}{\,\sqrt{2}-1\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{2}+1\,}{\,\sqrt{2}+1\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{2}+1\,}{\,(\sqrt{2}-1)(\sqrt{2}+1)\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{2}+1\,}{\,(\sqrt{2})^2-1^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{2}+1\,}{\,2-1\,}
\\[3pt]~~~&=&\sqrt{2}+1\end{eqnarray}\)
これより、
\(\begin{eqnarray}~~~x+y&=&(\sqrt{2}-1)+(\sqrt{2}+1)
\\[3pt]~~~&=&2\sqrt{2}~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\(\begin{eqnarray}~~~xy&=&(\sqrt{2}-1)(\sqrt{2}+1)
\\[3pt]~~~&=&(\sqrt{2})^2-1^2
\\[3pt]~~~&=&2-1
\\[3pt]~~~&=&1~~~\cdots {\small [\,2\,]}\end{eqnarray}\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~x^2+y^2&=&(x+y)^2-2xy
\\[3pt]~~~&=&(2\sqrt{2})^2-2 \cdot 1\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&4 \cdot 2-2
\\[3pt]~~~&=&8-2
\\[3pt]~~~&=&6\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~x^3+y^3&=&(x+y)^3-3xy(x+y)
\\[3pt]~~~&=&(2\sqrt{2})^3-3 \cdot 1 \cdot 2\sqrt{2}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&8 \cdot 2\sqrt{2}-6\sqrt{2}
\\[3pt]~~~&=&16\sqrt{2}-6\sqrt{2}
\\[3pt]~~~&=&10\sqrt{2}\end{eqnarray}\)
\\[3pt]~~~&=&(2\sqrt{2})^3-3 \cdot 1 \cdot 2\sqrt{2}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&8 \cdot 2\sqrt{2}-6\sqrt{2}
\\[3pt]~~~&=&16\sqrt{2}-6\sqrt{2}
\\[3pt]~~~&=&10\sqrt{2}\end{eqnarray}\)
※ 数式は横にスクロールできます。
\({\small (3)}~\)
\(\begin{eqnarray}~~~&&x^3y-x^2y^2+xy^3
\\[3pt]~~~&=&xy(x^2-xy+y^2)
\\[3pt]~~~&=&xy\{(x^2+y^2)-xy\}
\\[3pt]~~~&=&1 \cdot (6-1)\hspace{30pt}(\,∵~{\small [\,2\,]}~,~{\small (1)}\,)
\\[3pt]~~~&=&5\end{eqnarray}\)
問題アーカイブ02
問題アーカイブ02\(x=\displaystyle \frac{\,\sqrt{5}-1\,}{\,\sqrt{5}+1\,}~,~\)\(y=\displaystyle \frac{\,\sqrt{5}+1\,}{\,\sqrt{5}-1\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x^2+y^2\)
\({\small (2)}~\)\(x^2-y^2\)
\({\small (1)}~\)\(x^2+y^2\)
\({\small (2)}~\)\(x^2-y^2\)
数研出版｜数学Ⅰ[712] p.37 問題 12
\(x\) と \(y\) をそれぞれ有理化すると、
\(\begin{eqnarray}~~~x&=&\displaystyle \frac{\,\sqrt{5}-1\,}{\,\sqrt{5}+1\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{5}-1\,}{\,\sqrt{5}-1\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{5}-1)^2\,}{\,(\sqrt{5}+1)(\sqrt{5}-1)\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{5})^2-2\sqrt{5} \cdot 1+1^2\,}{\,(\sqrt{5})^2-1^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,5-2\sqrt{5}+1\,}{\,5-1\,}
\\[5pt]~~~&=&\displaystyle \frac{\,6-2\sqrt{5}\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,3-\sqrt{5}\,}{\,2\,}\end{eqnarray}\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,\sqrt{5}+1\,}{\,\sqrt{5}-1\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{5}+1\,}{\,\sqrt{5}+1\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{5}+1)^2\,}{\,(\sqrt{5}-1)(\sqrt{5}+1)\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{5})^2+2\sqrt{5} \cdot 1+1^2\,}{\,(\sqrt{5})^2-1^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,5+2\sqrt{5}+1\,}{\,5-1\,}
\\[5pt]~~~&=&\displaystyle \frac{\,6+2\sqrt{5}\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,3+\sqrt{5}\,}{\,2\,}\end{eqnarray}\)
これより、
\(\begin{eqnarray}~~~x+y&=&\displaystyle \frac{\,3-\sqrt{5}\,}{\,2\,}+\displaystyle \frac{\,3+\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(3-\sqrt{5})+(3+\sqrt{5})\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,6\,}{\,2\,}
\\[5pt]~~~&=&3~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\(\begin{eqnarray}~~~xy&=&\displaystyle \frac{\,3-\sqrt{5}\,}{\,2\,} \cdot \displaystyle \frac{\,3+\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(3-\sqrt{5})(3+\sqrt{5})\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,3^2-(\sqrt{5})^2\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,9-5\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,4\,}{\,4\,}
\\[5pt]~~~&=&1~~~\cdots {\small [\,2\,]}\end{eqnarray}\)
\(\begin{eqnarray}~~~x-y&=&\displaystyle \frac{\,3-\sqrt{5}\,}{\,2\,}-\displaystyle \frac{\,3+\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(3-\sqrt{5})-(3+\sqrt{5})\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,-2\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&-\sqrt{5}~~~\cdots {\small [\,3\,]}\end{eqnarray}\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~x^2+y^2&=&(x+y)^2-2xy
\\[3pt]~~~&=&3^2-2 \cdot 1\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&9-2
\\[3pt]~~~&=&7\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~x^2-y^2&=&(x+y)(x-y)
\\[3pt]~~~&=&3 \cdot (-\sqrt{5})\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,3\,]}\,)
\\[3pt]~~~&=&-3\sqrt{5}\end{eqnarray}\)
問題アーカイブ03
問題アーカイブ03\(x=\displaystyle \frac{\,1\,}{\,\sqrt{7}-\sqrt{5}\,}~,~\)\(y=\displaystyle \frac{\,1\,}{\,\sqrt{7}+\sqrt{5}\,}\) のとき、\(x^3+y^3\) の値を求めよ。
東京書籍｜Advanced数学Ⅰ[701] p.30 発展 問1
\(x\) と \(y\) をそれぞれ有理化すると、
\(\begin{eqnarray}~~~x&=&\displaystyle \frac{\,1\,}{\,\sqrt{7}-\sqrt{5}\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,\sqrt{7}+\sqrt{5}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,(\sqrt{7})^2-(\sqrt{5})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,7-5\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,2\,}\end{eqnarray}\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,1\,}{\,\sqrt{7}+\sqrt{5}\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,\sqrt{7}-\sqrt{5}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,(\sqrt{7})^2-(\sqrt{5})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,7-5\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,2\,}\end{eqnarray}\)
これより、
\(\begin{eqnarray}~~~x+y&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,2\,}+\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{7}+\sqrt{5})+(\sqrt{7}-\sqrt{5})\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2\sqrt{7}\,}{\,2\,}
\\[5pt]~~~&=&\sqrt{7}~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\(\begin{eqnarray}~~~xy&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,2\,} \cdot \displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{7})^2-(\sqrt{5})^2\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,7-5\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}~~~\cdots {\small [\,2\,]}\end{eqnarray}\)
\(\begin{eqnarray}~~~x^3+y^3&=&(x+y)^3-3xy(x+y)
\\[5pt]~~~&=&(\sqrt{7})^3-3 \cdot \displaystyle \frac{\,1\,}{\,2\,} \cdot \sqrt{7}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&7\sqrt{7}-\displaystyle \frac{\,3\sqrt{7}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,14\sqrt{7}-3\sqrt{7}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,11\sqrt{7}\,}{\,2\,}\end{eqnarray}\)
\\[5pt]~~~&=&(\sqrt{7})^3-3 \cdot \displaystyle \frac{\,1\,}{\,2\,} \cdot \sqrt{7}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&7\sqrt{7}-\displaystyle \frac{\,3\sqrt{7}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,14\sqrt{7}-3\sqrt{7}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,11\sqrt{7}\,}{\,2\,}\end{eqnarray}\)
※ 数式は横にスクロールできます。
問題アーカイブ04
問題アーカイブ04\(x=\displaystyle \frac{\,1\,}{\,\sqrt{5}-2\,}~,~\)\(y=\displaystyle \frac{\,1\,}{\,\sqrt{5}+2\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x^2-y^2\)
\({\small (2)}~\)\(\displaystyle \frac{\,x\,}{\,y\,}+\displaystyle \frac{\,y\,}{\,x\,}\)
\({\small (1)}~\)\(x^2-y^2\)
\({\small (2)}~\)\(\displaystyle \frac{\,x\,}{\,y\,}+\displaystyle \frac{\,y\,}{\,x\,}\)
東京書籍｜Advanced数学Ⅰ[701] p.32 問題 8
\(x\) と \(y\) をそれぞれ有理化すると、
\(\begin{eqnarray}~~~x&=&\displaystyle \frac{\,1\,}{\,\sqrt{5}-2\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{5}+2\,}{\,\sqrt{5}+2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{5}+2\,}{\,(\sqrt{5}-2)(\sqrt{5}+2)\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{5}+2\,}{\,(\sqrt{5})^2-2^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{5}+2\,}{\,5-4\,}
\\[3pt]~~~&=&\sqrt{5}+2\end{eqnarray}\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,1\,}{\,\sqrt{5}+2\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{5}-2\,}{\,\sqrt{5}-2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{5}-2\,}{\,(\sqrt{5}+2)(\sqrt{5}-2)\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{5}-2\,}{\,(\sqrt{5})^2-2^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{5}-2\,}{\,5-4\,}
\\[3pt]~~~&=&\sqrt{5}-2\end{eqnarray}\)
これより、
\(\begin{eqnarray}~~~x+y&=&(\sqrt{5}+2)+(\sqrt{5}-2)
\\[3pt]~~~&=&2\sqrt{5}~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\(\begin{eqnarray}~~~xy&=&(\sqrt{5}+2)(\sqrt{5}-2)
\\[3pt]~~~&=&(\sqrt{5})^2-2^2
\\[3pt]~~~&=&5-4
\\[3pt]~~~&=&1~~~\cdots {\small [\,2\,]}\end{eqnarray}\)
\(\begin{eqnarray}~~~x-y&=&(\sqrt{5}+2)-(\sqrt{5}-2)
\\[3pt]~~~&=&4~~~\cdots {\small [\,3\,]}\end{eqnarray}\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~x^2-y^2&=&(x+y)(x-y)
\\[3pt]~~~&=&2\sqrt{5} \cdot 4\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,3\,]}\,)
\\[3pt]~~~&=&8\sqrt{5}\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~\displaystyle \frac{\,x\,}{\,y\,}+\displaystyle \frac{\,y\,}{\,x\,}&=&\displaystyle \frac{\,x^2+y^2\,}{\,xy\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(x+y)^2-2xy\,}{\,xy\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(2\sqrt{5})^2-2 \cdot 1\,}{\,1\,}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&4 \cdot 5-2
\\[3pt]~~~&=&20-2
\\[3pt]~~~&=&18\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,(x+y)^2-2xy\,}{\,xy\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(2\sqrt{5})^2-2 \cdot 1\,}{\,1\,}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&4 \cdot 5-2
\\[3pt]~~~&=&20-2
\\[3pt]~~~&=&18\end{eqnarray}\)
※ 数式は横にスクロールできます。
