次の式の分母を有理化しよう。
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【解答】
\def\MA{\sqrt{2}} \def\MBK{} \def\MB{\sqrt{3}} \def\KA{\sqrt{6}} \def\KB{3} \newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{分母分子に}\colBX{mistyrose}{$\scriptsize\color{red}\MB$}\colMM{red}{をかける}\\ \dfrac{\MA}{\MBK\colBX{mistyrose}{$\MB$}} &= \dfrac{\MA\ \colBX{mistyrose}{$\times\MB$}}{\MBK\MB\ \colBX{mistyrose}{$\times\MB$}}\\ \\ &= \dfrac{\KA}{\KB} \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
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【解答】
\def\MA{1} \def\MBK{2} \def\MB{\sqrt{3}} \def\KA{\sqrt{3}} \def\KB{6} \newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{分母分子に}\colBX{mistyrose}{$\scriptsize\color{red}\MB$}\colMM{red}{をかける}\\ \dfrac{\MA}{\MBK\colBX{mistyrose}{$\MB$}} &= \dfrac{\MA\ \colBX{mistyrose}{$\times\MB$}}{\MBK\MB\ \colBX{mistyrose}{$\times\MB$}}\\ \\ &= \dfrac{\KA}{\KB} \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
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【解答】
\def\MA{6} \def\MBK{} \def\MB{\sqrt{3}} \def\KA{6\sqrt{3}} \def\KB{3} \def\Kotae{= 2\sqrt{3}} \newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \colMM{red}{分母分子に}\colBX{mistyrose}{$\scriptsize\color{red}\MB$}\colMM{red}{をかける}\\ \dfrac{\MA}{\MBK\colBX{mistyrose}{$\MB$}} &= \dfrac{\MA\ \colBX{mistyrose}{$\times\MB$}}{\MBK\MB\ \colBX{mistyrose}{$\times\MB$}}\\ \\ &= \dfrac{\KA}{\KB} \Kotae \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
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【解答】
\def\bunsi{1} \def\bunbo{\sqrt{5}+\sqrt{2}} \def\bunboy{\sqrt{5}-\sqrt{2}} \begin{align*} \dfrac{\bunsi}{\bunbo} &= \dfrac{1 \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}{\left(\bunbo\right) \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}\\ \\ &= \dfrac{\bunboy}{\sqrt{5}^2-\sqrt{2}^2}\\ \\ &= \dfrac{\bunboy}{5-2} = \dfrac{\bunboy}{3} \end{align*}
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【解答】
\def\bunsi{\sqrt{3}+1} \def\bunbo{\sqrt{3}-1} \def\bunboy{\sqrt{3}+1} \begin{align*} & \color{red}\scriptsize \Darr 分母分子にかける\\ \dfrac{\bunsi}{\bunbo} &= \dfrac{\left(\bunsi\right) \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}{\left(\bunbo\right) \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}\\ & \color{red}\scriptsize \searrow 分母1ヶ所符号を変えて \uparrow\\ &= \dfrac{\left(\bunboy\right)^2}{\sqrt{3}^2-1^2}\\ \\ &= \dfrac{\sqrt{3}^2+\sqrt{3} \times 2 + 1^2}{3-1}\\ \\ &= \dfrac{3+2\sqrt{3}+1}{2}\\ \\ &= \dfrac{4+2\sqrt{3}}{2}\\ \\ &= \dfrac{2\left(2+\sqrt{3}\right)}{2} = 2+\sqrt{3} \end{align*}
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【解答】
\def\bunsi{1} \def\bunbo{\sqrt{3}+\sqrt{2}} \def\bunboy{\sqrt{3}-\sqrt{2}} \begin{align*} \dfrac{\bunsi}{\bunbo} &= \dfrac{1 \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}{\left(\bunbo\right) \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}\\ \\ &= \dfrac{\bunboy}{\sqrt{3}^2-\sqrt{2}^2}\\ \\ &= \dfrac{\bunboy}{3-2}\\ \\ &= \dfrac{\bunboy}{1} = \bunboy \end{align*}
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【解答】
\def\bunsi{\sqrt{5}+\sqrt{2}} \def\bunbo{\sqrt{5}-\sqrt{2}} \def\bunboy{\sqrt{5}+\sqrt{2}} \begin{align*} & \color{red}\scriptsize \Darr 分母分子にかける\\ \dfrac{\bunsi}{\bunbo} &= \dfrac{\left(\bunsi\right) \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}{\left(\bunbo\right) \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}\\ & \color{red}\scriptsize \searrow 分母1ヶ所符号を変えて \uparrow\\ &= \dfrac{\left(\bunboy\right)^2}{\sqrt{5}^2-\sqrt{2}^2}\\ \\ &= \dfrac{\sqrt{5}^2+\sqrt{10} \times 2 + \sqrt{2}^2}{5-2}\\ \\ &= \dfrac{5+2\sqrt{10}+2}{3}\\ \\ &= \dfrac{7+2\sqrt{10}}{3} \end{align*}
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【解答】
\def\bunsi{\sqrt{3}} \def\bunbo{\sqrt{3}+1} \def\bunboy{\sqrt{3}-1} \begin{align*} \dfrac{\bunsi}{\bunbo} &= \dfrac{\bunsi \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}{\left(\bunbo\right) \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}\\ \\ &= \dfrac{\sqrt{3}\sqrt{3}-\sqrt{3}}{\sqrt{3}^2-1^2}\\ \\ &= \dfrac{3-\sqrt{3}}{3-1}\\ \\ &= \dfrac{3-\sqrt{3}}{2} \end{align*}
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【解答】
\def\bunsi{\sqrt{3}} \def\bunbo{\sqrt{3}+\sqrt{2}} \def\bunboy{\sqrt{3}-\sqrt{2}} \begin{align*} \dfrac{\bunsi}{\bunbo} &= \dfrac{\bunsi \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}{\left(\bunbo\right) \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}\\ \\ &= \dfrac{\sqrt{3}\sqrt{3}-\sqrt{3}\sqrt{2}}{\sqrt{3}^2-\sqrt{2}^2}\\ \\ &= \dfrac{3-\sqrt{6}}{3-2}\\ \\ &= \dfrac{3-\sqrt{6}}{1} = 3-\sqrt{6} \end{align*}
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【解答】
\def\bunsi{2-\sqrt{3}} \def\bunbo{2+\sqrt{3}} \def\bunboy{2-\sqrt{3}} \begin{align*} & \color{red}\scriptsize \Darr 分母分子にかける\\ \dfrac{\bunsi}{\bunbo} &= \dfrac{\left(\bunsi\right) \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}{\left(\bunbo\right) \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}\\ & \color{red}\scriptsize \searrow 分母1ヶ所符号を変えて \uparrow\\ &= \dfrac{\left(\bunboy\right)^2}{2^2-\sqrt{3}^2}\\ \\ &= \dfrac{2^2-2\sqrt{3} \times 2 + \sqrt{3}^2}{4-3}\\ \\ &= \dfrac{4-4\sqrt{3}+3}{1}\\ \\ &= 7-4\sqrt{3} \end{align*}
↓この問題へのリンクはこちら(右クリックで保存)
【解答】
\def\bunsi{1+\sqrt{5}} \def\bunbo{5-3\sqrt{5}} \def\bunboy{5+3\sqrt{5}} \newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \color{red}\scriptsize \Darr 分母分子にかける\\ \dfrac{\bunsi}{\bunbo} &= \dfrac{\left(\bunsi\right) \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}{\left(\bunbo\right) \times \left(\colorbox{mistyrose}{$\bunboy$}\right)}\\ & \color{red}\scriptsize \searrow 分母1ヶ所符号を変えて \uparrow\\ &= \dfrac{5+3\sqrt{5}+5\sqrt{5}+3 \cdot 5}{5^2-(3\sqrt{5})^2}\\ \\ &= \dfrac{\colBX{bisque}{$20+8\sqrt{5}$}}{25-9 \cdot 5}\\ & \colMM{orange}{ \Darr 因数分解}\\ &= \dfrac{\colBX{bisque}{$4(5+2\sqrt{5})$}}{25-45}\\ \\ &= \dfrac{4(5+2\sqrt{5})}{-20}\color{lightgray}= \dfrac{1(5+2\sqrt{5})}{-5}\\ & \colMM{green}{ \Darr 約分}\\ &= \dfrac{5+2\sqrt{5}}{-5}\\ & \colMM{violet}{ \Darr 分母のマイナスは,分数の前に出す!}\\ &= -\dfrac{5+2\sqrt{5}}{5} \end{align*}