$(\sqrt{5} + \sqrt{2} - 1)^2$ を計算してください。代数学式の展開根号計算2025/4/171. 問題の内容(5+2−1)2(\sqrt{5} + \sqrt{2} - 1)^2(5+2−1)2 を計算してください。2. 解き方の手順(5+2−1)2(\sqrt{5} + \sqrt{2} - 1)^2(5+2−1)2 を展開します。(5+2−1)2=(5+2−1)(5+2−1)(\sqrt{5} + \sqrt{2} - 1)^2 = (\sqrt{5} + \sqrt{2} - 1)(\sqrt{5} + \sqrt{2} - 1)(5+2−1)2=(5+2−1)(5+2−1)分配法則を用いて展開します。=(5)2+(5)(2)−5+(2)(5)+(2)2−2−5−2+1= (\sqrt{5})^2 + (\sqrt{5})(\sqrt{2}) - \sqrt{5} + (\sqrt{2})(\sqrt{5}) + (\sqrt{2})^2 - \sqrt{2} - \sqrt{5} - \sqrt{2} + 1=(5)2+(5)(2)−5+(2)(5)+(2)2−2−5−2+1=5+10−5+10+2−2−5−2+1= 5 + \sqrt{10} - \sqrt{5} + \sqrt{10} + 2 - \sqrt{2} - \sqrt{5} - \sqrt{2} + 1=5+10−5+10+2−2−5−2+1同類項をまとめます。=5+2+1+10+10−5−5−2−2= 5 + 2 + 1 + \sqrt{10} + \sqrt{10} - \sqrt{5} - \sqrt{5} - \sqrt{2} - \sqrt{2}=5+2+1+10+10−5−5−2−2=8+210−25−22= 8 + 2\sqrt{10} - 2\sqrt{5} - 2\sqrt{2}=8+210−25−22=8+2(10−5−2)= 8 + 2(\sqrt{10} - \sqrt{5} - \sqrt{2})=8+2(10−5−2)3. 最終的な答え8+210−25−228 + 2\sqrt{10} - 2\sqrt{5} - 2\sqrt{2}8+210−25−22