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問題アーカイブ01
問題アーカイブ01\(a~,~b\) は正の数とする。次の式を簡単にせよ。
\({\small (1)}~\)\((a^{\large \frac{1}{2}}-a^{\large -\frac{1}{2}})^2\)
\({\small (2)}~\)\((a^{\large \frac{1}{3}}-b^{\large -\frac{1}{3}})(a^{\large \frac{2}{3}}+a^{\large \frac{1}{3}}b^{\large -\frac{1}{3}}+b^{\large -\frac{2}{3}})\)
\({\small (1)}~\)\((a^{\large \frac{1}{2}}-a^{\large -\frac{1}{2}})^2\)
\({\small (2)}~\)\((a^{\large \frac{1}{3}}-b^{\large -\frac{1}{3}})(a^{\large \frac{2}{3}}+a^{\large \frac{1}{3}}b^{\large -\frac{1}{3}}+b^{\large -\frac{2}{3}})\)
数研出版｜数学Ⅱ[709] p.174 問題 2
\({\small (1)}~\)
展開の公式 \((x-y)^2=x^2-2xy+y^2\) より、
\(\begin{eqnarray}~~~&&(a^{\large \frac{1}{2}}-a^{\large -\frac{1}{2}})^2
\\[5pt]~~~&=&(a^{\large \frac{1}{2}})^2-2\cdot a^{\large \frac{1}{2}}\cdot a^{\large -\frac{1}{2}}+(a^{\large -\frac{1}{2}})^2
\\[5pt]~~~&=&a-2\cdot a^{\large \frac{1}{2}-\frac{1}{2}}+a^{-1}
\\[5pt]~~~&=&a-2\cdot a^0+\displaystyle\frac{\,1\,}{\,a\,}
\\[5pt]~~~&=&a-2+\displaystyle\frac{\,1\,}{\,a\,}\end{eqnarray}\)
\\[5pt]~~~&=&(a^{\large \frac{1}{2}})^2-2\cdot a^{\large \frac{1}{2}}\cdot a^{\large -\frac{1}{2}}+(a^{\large -\frac{1}{2}})^2
\\[5pt]~~~&=&a-2\cdot a^{\large \frac{1}{2}-\frac{1}{2}}+a^{-1}
\\[5pt]~~~&=&a-2\cdot a^0+\displaystyle\frac{\,1\,}{\,a\,}
\\[5pt]~~~&=&a-2+\displaystyle\frac{\,1\,}{\,a\,}\end{eqnarray}\)
\({\small (2)}~\)
展開の公式 \((x-y)(x^2+xy+y^2)=x^3-y^3\) より、
\(\begin{eqnarray}~~~&&(a^{\large \frac{1}{3}}-b^{\large -\frac{1}{3}})(a^{\large \frac{2}{3}}+a^{\large \frac{1}{3}}b^{\large -\frac{1}{3}}+b^{\large -\frac{2}{3}})
\\[5pt]~~~&=&(a^{\large \frac{1}{3}}-b^{\large -\frac{1}{3}})\left\{(a^{\large \frac{1}{3}})^2+a^{\large \frac{1}{3}}\cdot b^{\large -\frac{1}{3}}+(b^{\large -\frac{1}{3}})^2\right\}
\\[5pt]~~~&=&(a^{\large \frac{1}{3}})^3-(b^{\large -\frac{1}{3}})^3
\\[5pt]~~~&=&a-b^{-1}
\\[5pt]~~~&=&a-\displaystyle\frac{\,1\,}{\,b\,}\end{eqnarray}\)
\\[5pt]~~~&=&(a^{\large \frac{1}{3}}-b^{\large -\frac{1}{3}})\left\{(a^{\large \frac{1}{3}})^2+a^{\large \frac{1}{3}}\cdot b^{\large -\frac{1}{3}}+(b^{\large -\frac{1}{3}})^2\right\}
\\[5pt]~~~&=&(a^{\large \frac{1}{3}})^3-(b^{\large -\frac{1}{3}})^3
\\[5pt]~~~&=&a-b^{-1}
\\[5pt]~~~&=&a-\displaystyle\frac{\,1\,}{\,b\,}\end{eqnarray}\)
問題アーカイブ02
問題アーカイブ02次の式を計算せよ。
\({\small (1)}~\)\((\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}\,)(\sqrt[3]{3}-\sqrt[3]{2}\,)\)
\({\small (1)}~\)\((\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}\,)(\sqrt[3]{3}-\sqrt[3]{2}\,)\)
\({\small (2)}~\)\((2^{\large \frac{1}{2}}+2^{\large \frac{3}{4}}{\,\small\times\,}3^{\large \frac{1}{4}}+3^{\large \frac{1}{2}})(2^{\large \frac{1}{2}}-2^{\large \frac{3}{4}}{\,\small\times\,}3^{\large \frac{1}{4}}+3^{\large \frac{1}{2}})\)
数研出版｜数学Ⅱ[709] p.174 問題 3
\({\small (1)}~\)
\(\sqrt[3]{9}=\sqrt[3]{3^2}~,~\)\(\sqrt[3]{6}=\sqrt[3]{3\cdot 2}~,~\)\(\sqrt[3]{4}=\sqrt[3]{2^2}\) と変形すると、
\(\begin{eqnarray}~~~&&(\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}\,)(\sqrt[3]{3}-\sqrt[3]{2}\,)
\\[5pt]~~~&=&(\sqrt[3]{3^2}+\sqrt[3]{3\cdot 2}+\sqrt[3]{2^2}\,)(\sqrt[3]{3}-\sqrt[3]{2}\,)
\\[5pt]~~~&=&\left\{(\sqrt[3]{3}\,)^2+\sqrt[3]{3}\cdot\sqrt[3]{2}+(\sqrt[3]{2}\,)^2\right\}(\sqrt[3]{3}-\sqrt[3]{2}\,)\end{eqnarray}\)
\\[5pt]~~~&=&(\sqrt[3]{3^2}+\sqrt[3]{3\cdot 2}+\sqrt[3]{2^2}\,)(\sqrt[3]{3}-\sqrt[3]{2}\,)
\\[5pt]~~~&=&\left\{(\sqrt[3]{3}\,)^2+\sqrt[3]{3}\cdot\sqrt[3]{2}+(\sqrt[3]{2}\,)^2\right\}(\sqrt[3]{3}-\sqrt[3]{2}\,)\end{eqnarray}\)
展開の公式 \((x-y)(x^2+xy+y^2)=x^3-y^3\) より、
\(\begin{eqnarray}~~~&=&(\sqrt[3]{3}\,)^3-(\sqrt[3]{2}\,)^3
\\[5pt]~~~&=&3-2
\\[5pt]~~~&=&1\end{eqnarray}\)
\({\small (2)}~\)\(a=2^{\large \frac{1}{4}}~,~\)\(b=3^{\large \frac{1}{4}}\) とおくと、
\(\begin{eqnarray}~~~2^{\large \frac{1}{2}}&=&(2^{\large \frac{1}{4}})^2=a^2
\\[5pt]~~~2^{\large \frac{3}{4}}{\,\small\times\,}3^{\large \frac{1}{4}}&=&(2^{\large \frac{1}{4}})^3\cdot 3^{\large \frac{1}{4}}=a^3b
\\[5pt]~~~3^{\large \frac{1}{2}}&=&(3^{\large \frac{1}{4}})^2=b^2\end{eqnarray}\)
与式に代入すると、
\((a^2+a^3b+b^2)(a^2-a^3b+b^2)\)
展開の公式 \((x+y)(x-y)=x^2-y^2\) より、
\(\begin{eqnarray}~~~&&(a^2+a^3b+b^2)(a^2-a^3b+b^2)
\\[5pt]~~~&=&\left\{(a^2+b^2)+a^3b\right\}\left\{(a^2+b^2)-a^3b\right\}
\\[5pt]~~~&=&(a^2+b^2)^2-(a^3b)^2
\\[5pt]~~~&=&a^4+2a^2b^2+b^4-a^6b^2\end{eqnarray}\)
\\[5pt]~~~&=&\left\{(a^2+b^2)+a^3b\right\}\left\{(a^2+b^2)-a^3b\right\}
\\[5pt]~~~&=&(a^2+b^2)^2-(a^3b)^2
\\[5pt]~~~&=&a^4+2a^2b^2+b^4-a^6b^2\end{eqnarray}\)
\(a=2^{\large \frac{1}{4}}~,~\)\(b=3^{\large \frac{1}{4}}\) を戻すと、
\(\begin{eqnarray}~~~&=&(2^{\large \frac{1}{4}})^4+2(2^{\large \frac{1}{4}})^2(3^{\large \frac{1}{4}})^2+(3^{\large \frac{1}{4}})^4-(2^{\large \frac{1}{4}})^6(3^{\large \frac{1}{4}})^2
\\[5pt]~~~&=&2+2\cdot 2^{\large \frac{1}{2}}\cdot 3^{\large \frac{1}{2}}+3-2^{\large \frac{3}{2}}\cdot 3^{\large \frac{1}{2}}
\\[5pt]~~~&=&5+2\sqrt{6}-2\sqrt{6}
\\[5pt]~~~&=&5\end{eqnarray}\)
\\[5pt]~~~&=&2+2\cdot 2^{\large \frac{1}{2}}\cdot 3^{\large \frac{1}{2}}+3-2^{\large \frac{3}{2}}\cdot 3^{\large \frac{1}{2}}
\\[5pt]~~~&=&5+2\sqrt{6}-2\sqrt{6}
\\[5pt]~~~&=&5\end{eqnarray}\)
問題アーカイブ03
問題アーカイブ03次の式を計算せよ。
\({\small (1)}~\)\((\sqrt[4]{3}+\sqrt[4]{2}\,)(\sqrt[4]{3}-\sqrt[4]{2}\,)\)
\({\small (2)}~\)\((2^{\large \frac{1}{3}}-2^{\large -\frac{1}{3}})(2^{\large \frac{2}{3}}+1+2^{\large -\frac{2}{3}})\)
\({\small (1)}~\)\((\sqrt[4]{3}+\sqrt[4]{2}\,)(\sqrt[4]{3}-\sqrt[4]{2}\,)\)
\({\small (2)}~\)\((2^{\large \frac{1}{3}}-2^{\large -\frac{1}{3}})(2^{\large \frac{2}{3}}+1+2^{\large -\frac{2}{3}})\)
数研出版｜高等学校数学Ⅱ[710] p.165 問題 2(3)(4)
数研出版｜新編数学Ⅱ[711] p.160 補充問題 2(3)(4)
\({\small (1)}~\)展開の公式 \((x+y)(x-y)=x^2-y^2\) より、
\(\begin{eqnarray}~~~&&(\sqrt[4]{3}+\sqrt[4]{2}\,)(\sqrt[4]{3}-\sqrt[4]{2}\,)
\\[5pt]~~~&=&(\sqrt[4]{3}\,)^2-(\sqrt[4]{2}\,)^2
\\[5pt]~~~&=&\sqrt{3}-\sqrt{2}\end{eqnarray}\)
\({\small (2)}~\)展開の公式 \((x-y)(x^2+xy+y^2)=x^3-y^3\) より、
\(\begin{eqnarray}~~~&&(2^{\large \frac{1}{3}}-2^{\large -\frac{1}{3}})(2^{\large \frac{2}{3}}+1+2^{\large -\frac{2}{3}})
\\[5pt]~~~&=&(2^{\large \frac{1}{3}}-2^{\large -\frac{1}{3}})\left\{(2^{\large \frac{1}{3}})^2+2^{\large \frac{1}{3}}\cdot 2^{\large -\frac{1}{3}}+(2^{\large -\frac{1}{3}})^2\right\}
\\[5pt]~~~&=&(2^{\large \frac{1}{3}})^3-(2^{\large -\frac{1}{3}})^3
\\[5pt]~~~&=&2-2^{-1}
\\[5pt]~~~&=&2-\displaystyle\frac{\,1\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle\frac{\,3\,}{\,2\,}\end{eqnarray}\)
\\[5pt]~~~&=&(2^{\large \frac{1}{3}}-2^{\large -\frac{1}{3}})\left\{(2^{\large \frac{1}{3}})^2+2^{\large \frac{1}{3}}\cdot 2^{\large -\frac{1}{3}}+(2^{\large -\frac{1}{3}})^2\right\}
\\[5pt]~~~&=&(2^{\large \frac{1}{3}})^3-(2^{\large -\frac{1}{3}})^3
\\[5pt]~~~&=&2-2^{-1}
\\[5pt]~~~&=&2-\displaystyle\frac{\,1\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle\frac{\,3\,}{\,2\,}\end{eqnarray}\)
問題アーカイブ04
問題アーカイブ04次の計算をせよ。
\({\small (1)}~\)\((a^{\large \frac{1}{2}}+b^{\large \frac{1}{2}})(a^{\large \frac{1}{2}}-b^{\large \frac{1}{2}})\)
\({\small (2)}~\)\((a^{\large \frac{1}{3}}-b^{\large \frac{1}{3}})(a^{\large \frac{2}{3}}+a^{\large \frac{1}{3}}b^{\large \frac{1}{3}}+b^{\large \frac{2}{3}})\)
\({\small (1)}~\)\((a^{\large \frac{1}{2}}+b^{\large \frac{1}{2}})(a^{\large \frac{1}{2}}-b^{\large \frac{1}{2}})\)
\({\small (2)}~\)\((a^{\large \frac{1}{3}}-b^{\large \frac{1}{3}})(a^{\large \frac{2}{3}}+a^{\large \frac{1}{3}}b^{\large \frac{1}{3}}+b^{\large \frac{2}{3}})\)
東京書籍｜Advanced数学Ⅱ[701] p.170 問題 4(2)(3)
\({\small (1)}~\)展開の公式 \((x+y)(x-y)=x^2-y^2\) より、
\(\begin{eqnarray}~~~&&(a^{\large \frac{1}{2}}+b^{\large \frac{1}{2}})(a^{\large \frac{1}{2}}-b^{\large \frac{1}{2}})
\\[5pt]~~~&=&(a^{\large \frac{1}{2}})^2-(b^{\large \frac{1}{2}})^2
\\[5pt]~~~&=&a-b\end{eqnarray}\)
\\[5pt]~~~&=&(a^{\large \frac{1}{2}})^2-(b^{\large \frac{1}{2}})^2
\\[5pt]~~~&=&a-b\end{eqnarray}\)
\({\small (2)}~\)展開の公式 \((x-y)(x^2+xy+y^2)=x^3-y^3\) より、
\(\begin{eqnarray}~~~&&(a^{\large \frac{1}{3}}-b^{\large \frac{1}{3}})(a^{\large \frac{2}{3}}+a^{\large \frac{1}{3}}b^{\large \frac{1}{3}}+b^{\large \frac{2}{3}})
\\[5pt]~~~&=&(a^{\large \frac{1}{3}}-b^{\large \frac{1}{3}})\left\{(a^{\large \frac{1}{3}})^2+a^{\large \frac{1}{3}}\cdot b^{\large \frac{1}{3}}+(b^{\large \frac{1}{3}})^2\right\}
\\[5pt]~~~&=&(a^{\large \frac{1}{3}})^3-(b^{\large \frac{1}{3}})^3
\\[5pt]~~~&=&a-b\end{eqnarray}\)
\\[5pt]~~~&=&(a^{\large \frac{1}{3}}-b^{\large \frac{1}{3}})\left\{(a^{\large \frac{1}{3}})^2+a^{\large \frac{1}{3}}\cdot b^{\large \frac{1}{3}}+(b^{\large \frac{1}{3}})^2\right\}
\\[5pt]~~~&=&(a^{\large \frac{1}{3}})^3-(b^{\large \frac{1}{3}})^3
\\[5pt]~~~&=&a-b\end{eqnarray}\)
問題アーカイブ05
問題アーカイブ05次の式を簡単にせよ。
\({\small (1)}~\)\((a^{\large \frac{1}{2}}+a^{\large -\frac{1}{2}})^2\)
\({\small (2)}~\)\((a^{\large \frac{1}{2}}-b^{\large \frac{1}{2}})(a^{\large \frac{1}{2}}+b^{\large \frac{1}{2}})(a+b)\)
\({\small (3)}~\)\((a^{\large \frac{1}{3}}-a^{\large -\frac{1}{3}})(a^{\large \frac{2}{3}}+1+a^{\large -\frac{2}{3}})\)
\({\small (1)}~\)\((a^{\large \frac{1}{2}}+a^{\large -\frac{1}{2}})^2\)
\({\small (2)}~\)\((a^{\large \frac{1}{2}}-b^{\large \frac{1}{2}})(a^{\large \frac{1}{2}}+b^{\large \frac{1}{2}})(a+b)\)
\({\small (3)}~\)\((a^{\large \frac{1}{3}}-a^{\large -\frac{1}{3}})(a^{\large \frac{2}{3}}+1+a^{\large -\frac{2}{3}})\)
東京書籍｜Standard数学Ⅱ[702] p.194 Level Up 2
\({\small (1)}~\)展開の公式 \((x+y)^2=x^2+2xy+y^2\) より、
\(\begin{eqnarray}~~~&&(a^{\large \frac{1}{2}}+a^{\large -\frac{1}{2}})^2
\\[5pt]~~~&=&(a^{\large \frac{1}{2}})^2+2\cdot a^{\large \frac{1}{2}}\cdot a^{\large -\frac{1}{2}}+(a^{\large -\frac{1}{2}})^2
\\[5pt]~~~&=&a+2\cdot a^{\large \frac{1}{2}-\frac{1}{2}}+a^{-1}
\\[5pt]~~~&=&a+2\cdot a^0+\displaystyle\frac{\,1\,}{\,a\,}
\\[5pt]~~~&=&a+2+\displaystyle\frac{\,1\,}{\,a\,}\end{eqnarray}\)
\\[5pt]~~~&=&(a^{\large \frac{1}{2}})^2+2\cdot a^{\large \frac{1}{2}}\cdot a^{\large -\frac{1}{2}}+(a^{\large -\frac{1}{2}})^2
\\[5pt]~~~&=&a+2\cdot a^{\large \frac{1}{2}-\frac{1}{2}}+a^{-1}
\\[5pt]~~~&=&a+2\cdot a^0+\displaystyle\frac{\,1\,}{\,a\,}
\\[5pt]~~~&=&a+2+\displaystyle\frac{\,1\,}{\,a\,}\end{eqnarray}\)
\({\small (2)}~\)展開の公式 \((x+y)(x-y)=x^2-y^2\) より、
\(\begin{eqnarray}~~~&&(a^{\large \frac{1}{2}}-b^{\large \frac{1}{2}})(a^{\large \frac{1}{2}}+b^{\large \frac{1}{2}})(a+b)
\\[5pt]~~~&=&\left\{(a^{\large \frac{1}{2}})^2-(b^{\large \frac{1}{2}})^2\right\}(a+b)
\\[5pt]~~~&=&(a-b)(a+b)\end{eqnarray}\)
\\[5pt]~~~&=&\left\{(a^{\large \frac{1}{2}})^2-(b^{\large \frac{1}{2}})^2\right\}(a+b)
\\[5pt]~~~&=&(a-b)(a+b)\end{eqnarray}\)
さらに \((x+y)(x-y)=x^2-y^2\) より、
\(\begin{eqnarray}~~~&=&a^2-b^2\end{eqnarray}\)
\({\small (3)}~\)展開の公式 \((x-y)(x^2+xy+y^2)=x^3-y^3\) より、
\(\begin{eqnarray}~~~&&(a^{\large \frac{1}{3}}-a^{\large -\frac{1}{3}})(a^{\large \frac{2}{3}}+1+a^{\large -\frac{2}{3}})
\\[5pt]~~~&=&(a^{\large \frac{1}{3}}-a^{\large -\frac{1}{3}})\left\{(a^{\large \frac{1}{3}})^2+a^{\large \frac{1}{3}}\cdot a^{\large -\frac{1}{3}}+(a^{\large -\frac{1}{3}})^2\right\}
\\[5pt]~~~&=&(a^{\large \frac{1}{3}})^3-(a^{\large -\frac{1}{3}})^3
\\[5pt]~~~&=&a-a^{-1}
\\[5pt]~~~&=&a-\displaystyle\frac{\,1\,}{\,a\,}\end{eqnarray}\)
\\[5pt]~~~&=&(a^{\large \frac{1}{3}}-a^{\large -\frac{1}{3}})\left\{(a^{\large \frac{1}{3}})^2+a^{\large \frac{1}{3}}\cdot a^{\large -\frac{1}{3}}+(a^{\large -\frac{1}{3}})^2\right\}
\\[5pt]~~~&=&(a^{\large \frac{1}{3}})^3-(a^{\large -\frac{1}{3}})^3
\\[5pt]~~~&=&a-a^{-1}
\\[5pt]~~~&=&a-\displaystyle\frac{\,1\,}{\,a\,}\end{eqnarray}\)
