$\frac{2\sqrt{5} - 5\sqrt{2}}{\sqrt{5} - \sqrt{2}}$ の分母を有理化せよ。代数学分母の有理化平方根式の計算2025/6/221. 問題の内容25−525−2\frac{2\sqrt{5} - 5\sqrt{2}}{\sqrt{5} - \sqrt{2}}5−225−52 の分母を有理化せよ。2. 解き方の手順分母の有理化を行うために、分母の共役な複素数である 5+2\sqrt{5} + \sqrt{2}5+2 を分子と分母の両方に掛けます。25−525−2=(25−52)(5+2)(5−2)(5+2)\frac{2\sqrt{5} - 5\sqrt{2}}{\sqrt{5} - \sqrt{2}} = \frac{(2\sqrt{5} - 5\sqrt{2})(\sqrt{5} + \sqrt{2})}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})}5−225−52=(5−2)(5+2)(25−52)(5+2)分子を展開します。(25−52)(5+2)=255+252−525−522(2\sqrt{5} - 5\sqrt{2})(\sqrt{5} + \sqrt{2}) = 2\sqrt{5}\sqrt{5} + 2\sqrt{5}\sqrt{2} - 5\sqrt{2}\sqrt{5} - 5\sqrt{2}\sqrt{2}(25−52)(5+2)=255+252−525−522=2(5)+210−510−5(2)= 2(5) + 2\sqrt{10} - 5\sqrt{10} - 5(2)=2(5)+210−510−5(2)=10+210−510−10= 10 + 2\sqrt{10} - 5\sqrt{10} - 10=10+210−510−10=−310= -3\sqrt{10}=−310分母を展開します。(5−2)(5+2)=(5)2−(2)2(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2}) = (\sqrt{5})^2 - (\sqrt{2})^2(5−2)(5+2)=(5)2−(2)2=5−2= 5 - 2=5−2=3= 3=3したがって、(25−52)(5+2)(5−2)(5+2)=−3103\frac{(2\sqrt{5} - 5\sqrt{2})(\sqrt{5} + \sqrt{2})}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \frac{-3\sqrt{10}}{3}(5−2)(5+2)(25−52)(5+2)=3−310=−10= -\sqrt{10}=−103. 最終的な答え−10-\sqrt{10}−10