【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63

問題

次の式を計算せよ.

解説

• (1) $\dfrac{1}{1+\sqrt{2}-\sqrt{3}}$
  • $1^{2}+(\sqrt{2})^{2}=(\sqrt{3})^{2}$ を用いると,
    $\dfrac{1}{1+\sqrt{2}-\sqrt{3}}
    =\dfrac{1+\sqrt{2}+\sqrt{3}}{(1+\sqrt{2})^{2}-(\sqrt{3})^{2}}
    =\dfrac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$
    $=\dfrac{\sqrt{2}+2+\sqrt{6}}{4}
    =$$\dfrac{2+\sqrt{2}+\sqrt{6}}{4}$
• (2) $\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{\sqrt{5}+\sqrt{3}-\sqrt{2}}$
  • $(\sqrt{5})^{2}=(\sqrt{3})^{2}+(\sqrt{2})^{2}$ を用いると,
    $\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{\sqrt{5}+\sqrt{3}-\sqrt{2}}
    =\dfrac{(\sqrt{5}+\sqrt{3}+\sqrt{2})(\sqrt{5}-\sqrt{3}-\sqrt{2})}
    {(\sqrt{5}+\sqrt{3}-\sqrt{2})(\sqrt{5}-\sqrt{3}-\sqrt{2})}$
    分子 $=\sqrt{5}^{2}-(\sqrt{3}+\sqrt{2})^{2}=-2\sqrt{6}$
    分母 $=(\sqrt{5}+\sqrt{3})^{2}-(\sqrt{2})^{2}=6+2\sqrt{15}$
    より,
    $\dfrac{-2\sqrt{6}}{6+2\sqrt{15}}
    =\dfrac{-\sqrt{6}}{3+\sqrt{15}}
    =\dfrac{-\sqrt{6}(3-\sqrt{15})}{(3+\sqrt{15})(3-\sqrt{15})}
    =\dfrac{-\sqrt{6}(3-\sqrt{15})}{-6}$
    $=\dfrac{\sqrt{6}(3-\sqrt{15})}{6}
    =$$\dfrac{\sqrt{6}+\sqrt{15}}{3}$
• (3) $\dfrac{\sqrt{2}+\sqrt{5}+\sqrt{7}}{\sqrt{2}+\sqrt{5}-\sqrt{7}}
  +\dfrac{\sqrt{2}-\sqrt{5}+\sqrt{7}}{\sqrt{2}-\sqrt{5}-\sqrt{7}}$
  • $(\sqrt{2})^{2}+(\sqrt{5})^{2}=(\sqrt{7})^{2}$を用いると,
    $\dfrac{(\sqrt{2}+\sqrt{5}+\sqrt{7})^{2}}{2\sqrt{10}}
    +\dfrac{(\sqrt{2}-\sqrt{5}+\sqrt{7})^{2}}{-2\sqrt{10}}$
    $=\dfrac{2\cdot7+2\sqrt{10}+2\sqrt{14}}{2\sqrt{10}}
    -\dfrac{2\cdot7-2\sqrt{10}+2\sqrt{14}}{2\sqrt{10}}
    =\dfrac{4\sqrt{10}+2\sqrt{14}}{2\sqrt{10}}$
    $=2+\dfrac{\sqrt{14}}{\sqrt{10}}\cdot\sqrt{10}=$$2+\sqrt{14}$