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対称式を用いた式の値 で確認できます。
問題アーカイブ01
問題アーカイブ01\(x=\displaystyle \frac{\,1\,}{\,\sqrt{7}+\sqrt{5}\,}~,~\)\(y=\displaystyle \frac{\,1\,}{\,\sqrt{7}-\sqrt{5}\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x+y\)
\({\small (2)}~\)\(xy\)
\({\small (3)}~\)\(x^2+y^2\)
\({\small (4)}~\)\(x^2y+xy^2\)
\({\small (1)}~\)\(x+y\)
\({\small (2)}~\)\(xy\)
\({\small (3)}~\)\(x^2+y^2\)
\({\small (4)}~\)\(x^2y+xy^2\)
数研出版｜数学Ⅰ[712] p.34 練習31
\(x\) と \(y\) をそれぞれ有理化すると、
\(\begin{eqnarray}~~~x&=&\displaystyle \frac{\,1\,}{\,\sqrt{7}+\sqrt{5}\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,\sqrt{7}-\sqrt{5}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,(\sqrt{7})^2-(\sqrt{5})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,7-5\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,2\,}\end{eqnarray}\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,1\,}{\,\sqrt{7}-\sqrt{5}\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,\sqrt{7}+\sqrt{5}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,(\sqrt{7})^2-(\sqrt{5})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,7-5\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,2\,}\end{eqnarray}\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~x+y&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,2\,}+\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{7}-\sqrt{5})+(\sqrt{7}+\sqrt{5})\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2\sqrt{7}\,}{\,2\,}
\\[5pt]~~~&=&\sqrt{7}~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~xy&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,2\,} \cdot \displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{7})^2-(\sqrt{5})^2\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,7-5\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}~~~\cdots {\small [\,2\,]}\end{eqnarray}\)
\({\small (3)}~\)
\(\begin{eqnarray}~~~x^2+y^2&=&(x+y)^2-2xy
\\[5pt]~~~&=&(\sqrt{7})^2-2 \cdot \displaystyle \frac{\,1\,}{\,2\,}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&7-1
\\[3pt]~~~&=&6\end{eqnarray}\)
\\[5pt]~~~&=&(\sqrt{7})^2-2 \cdot \displaystyle \frac{\,1\,}{\,2\,}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&7-1
\\[3pt]~~~&=&6\end{eqnarray}\)
※ 数式は横にスクロールできます。
\({\small (4)}~\)
\(\begin{eqnarray}~~~x^2y+xy^2&=&xy(x+y)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,} \cdot \sqrt{7}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}\,}{\,2\,}\end{eqnarray}\)
問題アーカイブ02
問題アーカイブ02\(x=\displaystyle \frac{\,1\,}{\,\sqrt{7}-\sqrt{5}\,}~,~\)\(y=\displaystyle \frac{\,1\,}{\,\sqrt{7}+\sqrt{5}\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x+y~,~\)\(xy\)
\({\small (2)}~\)\(x^2+y^2\)
\({\small (1)}~\)\(x+y~,~\)\(xy\)
\({\small (2)}~\)\(x^2+y^2\)
数研出版｜高等学校数学Ⅰ[713] p.34 練習36
東京書籍｜Advanced数学Ⅰ[701] p.30 問18
\(x\) と \(y\) をそれぞれ有理化すると、
\(\begin{eqnarray}~~~x&=&\displaystyle \frac{\,1\,}{\,\sqrt{7}-\sqrt{5}\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,\sqrt{7}+\sqrt{5}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,(\sqrt{7})^2-(\sqrt{5})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,7-5\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,2\,}\end{eqnarray}\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,1\,}{\,\sqrt{7}+\sqrt{5}\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,\sqrt{7}-\sqrt{5}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,(\sqrt{7})^2-(\sqrt{5})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,7-5\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,2\,}\end{eqnarray}\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~x+y&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,2\,}+\displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{7}+\sqrt{5})+(\sqrt{7}-\sqrt{5})\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2\sqrt{7}\,}{\,2\,}
\\[5pt]~~~&=&\sqrt{7}~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\(\begin{eqnarray}~~~xy&=&\displaystyle \frac{\,\sqrt{7}+\sqrt{5}\,}{\,2\,} \cdot \displaystyle \frac{\,\sqrt{7}-\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{7})^2-(\sqrt{5})^2\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,7-5\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}~~~\cdots {\small [\,2\,]}\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~x^2+y^2&=&(x+y)^2-2xy
\\[5pt]~~~&=&(\sqrt{7})^2-2 \cdot \displaystyle \frac{\,1\,}{\,2\,}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&7-1
\\[3pt]~~~&=&6\end{eqnarray}\)
\\[5pt]~~~&=&(\sqrt{7})^2-2 \cdot \displaystyle \frac{\,1\,}{\,2\,}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&7-1
\\[3pt]~~~&=&6\end{eqnarray}\)
※ 数式は横にスクロールできます。
問題アーカイブ03
問題アーカイブ03\(x=\displaystyle \frac{\,1\,}{\,\sqrt{2}+1\,}~,~\)\(y=\displaystyle \frac{\,1\,}{\,\sqrt{2}-1\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x+y~,~\)\(xy\)
\({\small (2)}~\)\(x^2+y^2\)
\({\small (3)}~\)\(x^2y+xy^2\)
\({\small (1)}~\)\(x+y~,~\)\(xy\)
\({\small (2)}~\)\(x^2+y^2\)
\({\small (3)}~\)\(x^2y+xy^2\)
数研出版｜高等学校数学Ⅰ[713] p.34 練習37
\(x\) と \(y\) をそれぞれ有理化すると、
\(\begin{eqnarray}~~~x&=&\displaystyle \frac{\,1\,}{\,\sqrt{2}+1\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{2}-1\,}{\,\sqrt{2}-1\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{2}-1\,}{\,(\sqrt{2}+1)(\sqrt{2}-1)\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{2}-1\,}{\,(\sqrt{2})^2-1^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{2}-1\,}{\,2-1\,}
\\[3pt]~~~&=&\sqrt{2}-1\end{eqnarray}\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,1\,}{\,\sqrt{2}-1\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{2}+1\,}{\,\sqrt{2}+1\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{2}+1\,}{\,(\sqrt{2}-1)(\sqrt{2}+1)\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{2}+1\,}{\,(\sqrt{2})^2-1^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\sqrt{2}+1\,}{\,2-1\,}
\\[3pt]~~~&=&\sqrt{2}+1\end{eqnarray}\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~x+y&=&(\sqrt{2}-1)+(\sqrt{2}+1)
\\[3pt]~~~&=&2\sqrt{2}~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\(\begin{eqnarray}~~~xy&=&(\sqrt{2}-1)(\sqrt{2}+1)
\\[3pt]~~~&=&(\sqrt{2})^2-1^2
\\[3pt]~~~&=&2-1
\\[3pt]~~~&=&1~~~\cdots {\small [\,2\,]}\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~x^2+y^2&=&(x+y)^2-2xy
\\[3pt]~~~&=&(2\sqrt{2})^2-2 \cdot 1\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&4 \cdot 2-2
\\[3pt]~~~&=&8-2
\\[3pt]~~~&=&6\end{eqnarray}\)
\({\small (3)}~\)
\(\begin{eqnarray}~~~x^2y+xy^2&=&xy(x+y)
\\[3pt]~~~&=&1 \cdot 2\sqrt{2}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&2\sqrt{2}\end{eqnarray}\)
問題アーカイブ04
問題アーカイブ04\(x=\displaystyle \frac{\,\sqrt{5}-\sqrt{3}\,}{\,\sqrt{5}+\sqrt{3}\,}~,~\)\(y=\displaystyle \frac{\,\sqrt{5}+\sqrt{3}\,}{\,\sqrt{5}-\sqrt{3}\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x^2+y^2\)
\({\small (2)}~\)\(x^3y+xy^3\)
\({\small (3)}~\)\(\displaystyle \frac{\,x\,}{\,y\,}+\displaystyle \frac{\,y\,}{\,x\,}\)
\({\small (1)}~\)\(x^2+y^2\)
\({\small (2)}~\)\(x^3y+xy^3\)
\({\small (3)}~\)\(\displaystyle \frac{\,x\,}{\,y\,}+\displaystyle \frac{\,y\,}{\,x\,}\)
数研出版｜高等学校数学Ⅰ[713] p.36 問題 10
\(x\) と \(y\) をそれぞれ有理化すると、
\(\begin{eqnarray}~~~x&=&\displaystyle \frac{\,\sqrt{5}-\sqrt{3}\,}{\,\sqrt{5}+\sqrt{3}\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{5}-\sqrt{3}\,}{\,\sqrt{5}-\sqrt{3}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{5}-\sqrt{3})^2\,}{\,(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{5})^2-2\sqrt{5}\cdot\sqrt{3}+(\sqrt{3})^2\,}{\,(\sqrt{5})^2-(\sqrt{3})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,5-2\sqrt{15}+3\,}{\,5-3\,}
\\[5pt]~~~&=&\displaystyle \frac{\,8-2\sqrt{15}\,}{\,2\,}
\\[5pt]~~~&=&4-\sqrt{15}\end{eqnarray}\)
\(\begin{eqnarray}~~~y&=&\displaystyle \frac{\,\sqrt{5}+\sqrt{3}\,}{\,\sqrt{5}-\sqrt{3}\,}{\, \small \times \,}\displaystyle \frac{\,\sqrt{5}+\sqrt{3}\,}{\,\sqrt{5}+\sqrt{3}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{5}+\sqrt{3})^2\,}{\,(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{5})^2+2\sqrt{5}\cdot\sqrt{3}+(\sqrt{3})^2\,}{\,(\sqrt{5})^2-(\sqrt{3})^2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,5+2\sqrt{15}+3\,}{\,5-3\,}
\\[5pt]~~~&=&\displaystyle \frac{\,8+2\sqrt{15}\,}{\,2\,}
\\[5pt]~~~&=&4+\sqrt{15}\end{eqnarray}\)
これより、
\(\begin{eqnarray}~~~x+y&=&(4-\sqrt{15})+(4+\sqrt{15})
\\[3pt]~~~&=&8~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\(\begin{eqnarray}~~~xy&=&(4-\sqrt{15})(4+\sqrt{15})
\\[3pt]~~~&=&4^2-(\sqrt{15})^2
\\[3pt]~~~&=&16-15
\\[3pt]~~~&=&1~~~\cdots {\small [\,2\,]}\end{eqnarray}\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~x^2+y^2&=&(x+y)^2-2xy
\\[3pt]~~~&=&8^2-2 \cdot 1\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&64-2
\\[3pt]~~~&=&62\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~x^3y+xy^3&=&xy(x^2+y^2)
\\[3pt]~~~&=&1 \cdot 62\hspace{30pt}(\,∵~{\small [\,2\,]}~,~{\small (1)}\,)
\\[3pt]~~~&=&62\end{eqnarray}\)
\({\small (3)}~\)
\(\begin{eqnarray}~~~\displaystyle \frac{\,x\,}{\,y\,}+\displaystyle \frac{\,y\,}{\,x\,}&=&\displaystyle \frac{\,x^2+y^2\,}{\,xy\,}
\\[5pt]~~~&=&\displaystyle \frac{\,62\,}{\,1\,}\hspace{30pt}(\,∵~{\small [\,2\,]}~,~{\small (1)}\,)
\\[5pt]~~~&=&62\end{eqnarray}\)
問題アーカイブ05
問題アーカイブ05\(x=\displaystyle \frac{\,\sqrt{3}+\sqrt{5}\,}{\,2\,}~,~\)\(y=\displaystyle \frac{\,\sqrt{3}-\sqrt{5}\,}{\,2\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x+y\)
\({\small (2)}~\)\(xy\)
\({\small (3)}~\)\(x^2+y^2\)
\({\small (4)}~\)\(x^3y+xy^3\)
\({\small (1)}~\)\(x+y\)
\({\small (2)}~\)\(xy\)
\({\small (3)}~\)\(x^2+y^2\)
\({\small (4)}~\)\(x^3y+xy^3\)
数研出版｜新編数学Ⅰ[714] p.37 補充問題 5
\({\small (1)}~\)
\(\begin{eqnarray}~~~x+y&=&\displaystyle \frac{\,\sqrt{3}+\sqrt{5}\,}{\,2\,}+\displaystyle \frac{\,\sqrt{3}-\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{3}+\sqrt{5})+(\sqrt{3}-\sqrt{5})\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2\sqrt{3}\,}{\,2\,}
\\[5pt]~~~&=&\sqrt{3}~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~xy&=&\displaystyle \frac{\,\sqrt{3}+\sqrt{5}\,}{\,2\,} \cdot \displaystyle \frac{\,\sqrt{3}-\sqrt{5}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{3}+\sqrt{5})(\sqrt{3}-\sqrt{5})\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(\sqrt{3})^2-(\sqrt{5})^2\,}{\,4\,}
\\[5pt]~~~&=&\displaystyle \frac{\,3-5\,}{\,4\,}
\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,}~~~\cdots {\small [\,2\,]}\end{eqnarray}\)
\({\small (3)}~\)
\(\begin{eqnarray}~~~x^2+y^2&=&(x+y)^2-2xy
\\[5pt]~~~&=&(\sqrt{3})^2-2 \cdot \left(-\displaystyle \frac{\,1\,}{\,2\,}\right)\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&3+1
\\[3pt]~~~&=&4\end{eqnarray}\)
\\[5pt]~~~&=&(\sqrt{3})^2-2 \cdot \left(-\displaystyle \frac{\,1\,}{\,2\,}\right)\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&3+1
\\[3pt]~~~&=&4\end{eqnarray}\)
※ 数式は横にスクロールできます。
\({\small (4)}~\)
\(\begin{eqnarray}~~~x^3y+xy^3&=&xy(x^2+y^2)
\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,} \cdot 4\hspace{30pt}(\,∵~{\small [\,2\,]}~,~{\small (3)}\,)
\\[5pt]~~~&=&-2\end{eqnarray}\)
\\[5pt]~~~&=&-\displaystyle \frac{\,1\,}{\,2\,} \cdot 4\hspace{30pt}(\,∵~{\small [\,2\,]}~,~{\small (3)}\,)
\\[5pt]~~~&=&-2\end{eqnarray}\)
※ 数式は横にスクロールできます。
問題アーカイブ06
問題アーカイブ06\(x=\displaystyle \frac{\,1\,}{\,\sqrt{5}+2\,}~,~\)\(y=\displaystyle \frac{\,1\,}{\,\sqrt{5}-2\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~\)\(x^2+xy+y^2\)
\({\small (2)}~\)\(\displaystyle \frac{\,y\,}{\,x\,}+\displaystyle \frac{\,x\,}{\,y\,}\)
\({\small (1)}~\)\(x^2+xy+y^2\)
\({\small (2)}~\)\(\displaystyle \frac{\,y\,}{\,x\,}+\displaystyle \frac{\,x\,}{\,y\,}\)
東京書籍｜Standard数学Ⅰ[702] p.35 Training 10
\(x\) 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、
\(\begin{eqnarray}~~~x+y&=&(\sqrt{5}-2)+(\sqrt{5}+2)
\\[3pt]~~~&=&2\sqrt{5}~~~\cdots {\small [\,1\,]}\end{eqnarray}\)
\(\begin{eqnarray}~~~xy&=&(\sqrt{5}-2)(\sqrt{5}+2)
\\[3pt]~~~&=&(\sqrt{5})^2-2^2
\\[3pt]~~~&=&5-4
\\[3pt]~~~&=&1~~~\cdots {\small [\,2\,]}\end{eqnarray}\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~x^2+xy+y^2&=&(x^2+2xy+y^2)-xy
\\[3pt]~~~&=&(x+y)^2-xy
\\[3pt]~~~&=&(2\sqrt{5})^2-1\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&4 \cdot 5-1
\\[3pt]~~~&=&20-1
\\[3pt]~~~&=&19\end{eqnarray}\)
\\[3pt]~~~&=&(x+y)^2-xy
\\[3pt]~~~&=&(2\sqrt{5})^2-1\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&4 \cdot 5-1
\\[3pt]~~~&=&20-1
\\[3pt]~~~&=&19\end{eqnarray}\)
※ 数式は横にスクロールできます。
\({\small (2)}~\)
\(\begin{eqnarray}~~~\displaystyle \frac{\,y\,}{\,x\,}+\displaystyle \frac{\,x\,}{\,y\,}&=&\displaystyle \frac{\,x^2+y^2\,}{\,xy\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(x+y)^2-2xy\,}{\,xy\,}
\\[5pt]~~~&=&\displaystyle \frac{\,(2\sqrt{5})^2-2 \cdot 1\,}{\,1\,}\hspace{30pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&4 \cdot 5-2
\\[3pt]~~~&=&20-2
\\[3pt]~~~&=&18\end{eqnarray}\)
\\[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。
