## 問題の回答
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1. 問題の内容
与えられた6つの式をそれぞれ簡単にせよ。式はそれぞれ以下の通り。
1. $(2\sqrt{3}+5)(\sqrt{3}-1)$
2. $(\sqrt{5}+5\sqrt{2})(2\sqrt{5}-\sqrt{2})$
3. $(\sqrt{3}+\sqrt{2})(2\sqrt{3}-3\sqrt{2})$
4. $(4-2\sqrt{3})(4+\sqrt{12})$
5. $(\sqrt{3}+\sqrt{15})(\sqrt{60}-\sqrt{12})$
6. $(\sqrt{2}-\sqrt{3}+\sqrt{5})(\sqrt{2}-\sqrt{3}-\sqrt{5})$
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2. 解き方の手順
各式をそれぞれ展開し、整理していく。根号の中身を簡単にできる場合は簡単にする。
1. $(2\sqrt{3}+5)(\sqrt{3}-1) = 2\sqrt{3}\cdot\sqrt{3} - 2\sqrt{3} + 5\sqrt{3} - 5 = 6 + 3\sqrt{3} - 5 = 1 + 3\sqrt{3}$
2. $(\sqrt{5}+5\sqrt{2})(2\sqrt{5}-\sqrt{2}) = \sqrt{5}\cdot 2\sqrt{5} - \sqrt{5}\cdot\sqrt{2} + 5\sqrt{2}\cdot 2\sqrt{5} - 5\sqrt{2}\cdot\sqrt{2} = 10 - \sqrt{10} + 10\sqrt{10} - 10 = 9\sqrt{10}$
3. $(\sqrt{3}+\sqrt{2})(2\sqrt{3}-3\sqrt{2}) = \sqrt{3}\cdot 2\sqrt{3} - \sqrt{3}\cdot 3\sqrt{2} + \sqrt{2}\cdot 2\sqrt{3} - \sqrt{2}\cdot 3\sqrt{2} = 6 - 3\sqrt{6} + 2\sqrt{6} - 6 = -\sqrt{6}$
4. $(4-2\sqrt{3})(4+\sqrt{12}) = (4-2\sqrt{3})(4+2\sqrt{3}) = 4^2 - (2\sqrt{3})^2 = 16 - 4\cdot 3 = 16 - 12 = 4$
5. $(\sqrt{3}+\sqrt{15})(\sqrt{60}-\sqrt{12}) = (\sqrt{3}+\sqrt{15})(2\sqrt{15}-2\sqrt{3}) = \sqrt{3}\cdot 2\sqrt{15} - \sqrt{3}\cdot 2\sqrt{3} + \sqrt{15}\cdot 2\sqrt{15} - \sqrt{15}\cdot 2\sqrt{3} = 2\sqrt{45} - 6 + 30 - 2\sqrt{45} = 24$
6. $(\sqrt{2}-\sqrt{3}+\sqrt{5})(\sqrt{2}-\sqrt{3}-\sqrt{5}) = ((\sqrt{2}-\sqrt{3})+\sqrt{5})((\sqrt{2}-\sqrt{3})-\sqrt{5}) = (\sqrt{2}-\sqrt{3})^2 - (\sqrt{5})^2 = 2 - 2\sqrt{6} + 3 - 5 = -2\sqrt{6}$
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