【高校数学I】数と式を完全攻略～基本から応用までを徹底解説します！～実数-問題55

問題

次の式の分母を有理化せよ.

解説

• (1) $\dfrac{2}{\sqrt{5}}$
  • $\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}=$$\dfrac{2\sqrt{5}}{5}$
• (2) $\dfrac{4}{3\sqrt{8}}$
  • $\sqrt{8}=2\sqrt{2}$ より,
    $\dfrac{4}{3\sqrt{8}}=\dfrac{4}{3\times2\sqrt{2}}=\dfrac{2}{3\sqrt{2}}=\dfrac{2\sqrt{2}}{6}=$$\dfrac{\sqrt{2}}{3}$
• (3) $\dfrac{1}{\sqrt{2}+1}$
  • 分母に $\sqrt{2}-1$ をかけて,
    $\dfrac{1}{\sqrt{2}+1}\cdot\dfrac{\sqrt{2}-1}{\sqrt{2}-1}
    =\dfrac{\sqrt{2}-1}{2-1}=$$\sqrt{2}-1$
• (4) $\dfrac{1}{\sqrt{6}-\sqrt{3}}$
  • 分母に $\sqrt{6}+\sqrt{3}$ をかけて,
    $
    \dfrac{1}{\sqrt{6}-\sqrt{3}}\cdot\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}
    =\dfrac{\sqrt{6}+\sqrt{3}}{6-3}
    =$$\dfrac{\sqrt{6}+\sqrt{3}}{3}$
• (5) $\dfrac{2+\sqrt{3}}{2-\sqrt{3}}$
  • 分母に $2+\sqrt{3}$ をかけて,
    $
    \dfrac{2+\sqrt{3}}{2-\sqrt{3}}\cdot\dfrac{2+\sqrt{3}}{2+\sqrt{3}}
    =\dfrac{(2+\sqrt{3})^{2}}{4-3}
    =\dfrac{7+4\sqrt{3}}{1}
    =$$7+4\sqrt{3}$
• (6) $\dfrac{\sqrt{3}-1}{2\sqrt{3}-5}$
  • 分母に$2\sqrt{3}+5$をかけて,\\
    $
    \dfrac{\sqrt{3}-1}{2\sqrt{3}-5}\cdot\dfrac{2\sqrt{3}+5}{2\sqrt{3}+5}
    =\dfrac{(6+5\sqrt{3}-2\sqrt{3}-5)}{(2\sqrt{3})^{2}-5^{2}}
    =\dfrac{3\sqrt{3}+1}{12-25}
    =$$-\dfrac{1+3\sqrt{3}}{13}$
• (7) $\dfrac{2\sqrt{2}-\sqrt{3}}{\sqrt{3}+\sqrt{2}}$
  • 分母に $\sqrt{3}-\sqrt{2}$ をかけて,\\
    $
    \dfrac{2\sqrt{2}-\sqrt{3}}{\sqrt{3}+\sqrt{2}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}
    =\dfrac{(2\sqrt{2}-\sqrt{3})(\sqrt{3}-\sqrt{2})}{3-2}
    =2\sqrt{6}-4-3+\sqrt{6}
    =$$-7+3\sqrt{6}$