根号を含む式の計算での分母の有理化(2)

根号を含む式の計算での分母の有理化

分母が無理数では、計算が面倒になる事が多いので、有理化して計算する。

例題

$(1)\frac{3\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{2}}{2\sqrt{3}}$
$(2)\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}$

例題解答

$(1)\frac{3\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{2}}{2\sqrt{3}}\\
=\frac{3\sqrt{6}}{2}+\frac{\sqrt{6}}{6}\\
=\frac{9\sqrt{6}}{6}+\frac{\sqrt{6}}{6}\\
=\frac{5\sqrt{6}}{3}$

$(2)\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\\
=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}\\
=\sqrt{3}-1$

問題

$(1)\frac{3\sqrt{2}}{2\sqrt{3}}-\frac{\sqrt{3}}{3\sqrt{2}}+\frac{1}{2\sqrt{6}}$
$(2)\frac{8}{3-\sqrt{5}}-\frac{2}{2+\sqrt{5}}$
$(3)\frac{\sqrt{5}}{\sqrt{3}+1}-\frac{\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
$(4)\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}+\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

問題解答

$(1)\frac{3\sqrt{2}}{2\sqrt{3}}-\frac{\sqrt{3}}{3\sqrt{2}}+\frac{1}{2\sqrt{6}}\\
=\frac{6\sqrt{6}}{12}-\frac{2\sqrt{6}}{12}+\frac{\sqrt{6}}{12}\\
=\frac{5\sqrt{6}}{12}$

$(2)\frac{8}{3-\sqrt{5}}-\frac{2}{2+\sqrt{5}}\\
=\frac{8(3+\sqrt{5})}{4}-\frac{2(\sqrt{5}-2}{1}\\
=6+2\sqrt{5}-2\sqrt{5}+4\\
=10$

$(3)\frac{\sqrt{5}}{\sqrt{3}+1}-\frac{\sqrt{3}}{\sqrt{5}+\sqrt{3}}\\
=\frac{\sqrt{5}(\sqrt{3}-1)}{2}-\frac{\sqrt{3}(\sqrt{5}-\sqrt{3})}{2}\\
=\frac{\sqrt{15}-\sqrt{5}}{2}-\frac{\sqrt{15}-3}{2}\\
=\frac{3-\sqrt{5}}{2}$

$(4)\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}+\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\
=(\sqrt{3}-\sqrt{2})^2+(\sqrt{3}+\sqrt{2})^2\\
=3-2\sqrt{6}+2+3+2\sqrt{6}+2\\
=10$

問題

次の式(1)(2)を有理化せよ。また、(3)(4)の計算をせよ。
$(1)\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}$
$(2)\frac{\sqrt{5}+\sqrt{3}}{2\sqrt{5}-\sqrt{3}}$
$(3)\frac{3-\sqrt{6}}{3+\sqrt{6}}+\frac{\sqrt{2}}{3-\sqrt{6}}$
$(4)\sqrt{2}=1.414,\sqrt{3}=1.732とするとき、\\
\frac{3}{\sqrt{48}},\frac{2\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

問題解答

$(1)\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}\\
=\frac{7-2\sqrt{21}+3}{4}\\
=\frac{5-\sqrt{21}}{2}$

$(2)\frac{\sqrt{5}+\sqrt{3}}{2\sqrt{5}-\sqrt{3}}\\
=\frac{(\sqrt{5}+\sqrt{3})(2\sqrt{5}+\sqrt{3})}{17}\\
=\frac{13+3\sqrt{15}}{17}$

$(3)\frac{3-\sqrt{6}}{3+\sqrt{6}}+\frac{\sqrt{2}}{3-\sqrt{6}}\\
=\frac{(3-\sqrt{6})^2+\sqrt{2}(3+\sqrt{6})}{3}\\
=\frac{15-6\sqrt{6}+3\sqrt{2}+2\sqrt{3})}{3}$

$\frac{3}{\sqrt{48}}\\
=\frac{3}{4\sqrt{3}}\\
=\frac{\sqrt{3}}{4}\\
=0.433$

$\frac{2\sqrt{2}}{\sqrt{6}+\sqrt{2}}\\
=\frac{2\sqrt{2}(\sqrt{6}-\sqrt{2})}{4}\\
=\frac{4\sqrt{3}-4}{4}\\
=0.732$

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