$\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ を計算し、分母を有理化してください。算数分母の有理化平方根計算2025/5/271. 問題の内容5+25−2\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}5−25+2 を計算し、分母を有理化してください。2. 解き方の手順分母を有理化するために、分母の共役な複素数である (5+2)(\sqrt{5}+\sqrt{2})(5+2) を分子と分母に掛けます。5+25−2=(5+2)(5+2)(5−2)(5+2)\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}} = \frac{(\sqrt{5}+\sqrt{2})(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}5−25+2=(5−2)(5+2)(5+2)(5+2)分子を展開します:(5+2)(5+2)=(5)2+2(5)(2)+(2)2=5+210+2=7+210(\sqrt{5}+\sqrt{2})(\sqrt{5}+\sqrt{2}) = (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{2}) + (\sqrt{2})^2 = 5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10}(5+2)(5+2)=(5)2+2(5)(2)+(2)2=5+210+2=7+210分母を展開します:(5−2)(5+2)=(5)2−(2)2=5−2=3(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2}) = (\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3(5−2)(5+2)=(5)2−(2)2=5−2=3したがって、(5+2)(5+2)(5−2)(5+2)=7+2103\frac{(\sqrt{5}+\sqrt{2})(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})} = \frac{7 + 2\sqrt{10}}{3}(5−2)(5+2)(5+2)(5+2)=37+2103. 最終的な答え7+2103\frac{7 + 2\sqrt{10}}{3}37+210