次の計算をしなさい。 (1) $(\sqrt{3}+1)(\sqrt{3}+2)$ (2) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$代数学展開平方根式の計算有理化2025/5/181. 問題の内容次の計算をしなさい。(1) (3+1)(3+2)(\sqrt{3}+1)(\sqrt{3}+2)(3+1)(3+2)(2) (5−2)(5+2)(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})(5−2)(5+2)2. 解き方の手順(1) (3+1)(3+2)(\sqrt{3}+1)(\sqrt{3}+2)(3+1)(3+2) を展開します。(3+1)(3+2)=3×3+3×2+1×3+1×2(\sqrt{3}+1)(\sqrt{3}+2) = \sqrt{3} \times \sqrt{3} + \sqrt{3} \times 2 + 1 \times \sqrt{3} + 1 \times 2(3+1)(3+2)=3×3+3×2+1×3+1×2=3+23+3+2= 3 + 2\sqrt{3} + \sqrt{3} + 2=3+23+3+2=5+33= 5 + 3\sqrt{3}=5+33(2) (5−2)(5+2)(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})(5−2)(5+2) を展開します。これは和と差の積の公式 (a−b)(a+b)=a2−b2 (a-b)(a+b) = a^2 - b^2 (a−b)(a+b)=a2−b2 を利用できます。(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=33. 最終的な答え(1) 5+335 + 3\sqrt{3}5+33(2) 333