問題
$\sqrt{2}=1.4142,\ \sqrt{3}=1.7321$ とするとき, 分母の有理化を利用して, 次の値を求めよ.
解説
- (1) $\dfrac{10}{\sqrt{3}+\sqrt{2}}$
- $=\dfrac{10(\sqrt{3}-\sqrt{2})}{3-2}=10(\sqrt{3}-\sqrt{2})
\approx $$3.179$
- $=\dfrac{10(\sqrt{3}-\sqrt{2})}{3-2}=10(\sqrt{3}-\sqrt{2})
- (2) $\dfrac{1}{\sqrt{12}-\sqrt{2}}$
- $=\dfrac{2\sqrt{3}+\sqrt{2}}{(2\sqrt{3})^{2}-(\sqrt{2})^{2}}
=\dfrac{1}{5}\sqrt{3}+\dfrac{1}{10}\sqrt{2}
\approx $$0.48784$
- $=\dfrac{2\sqrt{3}+\sqrt{2}}{(2\sqrt{3})^{2}-(\sqrt{2})^{2}}
