与えられた5つの根号計算の問題を解く。 (1) $\sqrt{10} \times \sqrt{2}$ (2) $\sqrt{21} \div \sqrt{7}$ (3) $2\sqrt{6} - 9\sqrt{6}$ (4) $-\sqrt{27} + \sqrt{75}$ (5) $(\sqrt{2}+1)^2$

1. 問題の内容

与えられた5つの根号計算の問題を解く。

(1) $\sqrt{10} \times \sqrt{2}$

(2) $\sqrt{21} \div \sqrt{7}$

(3) $2\sqrt{6} - 9\sqrt{6}$

(4) $-\sqrt{27} + \sqrt{75}$

(5) $(\sqrt{2}+1)^2$

2. 解き方の手順

(1) $\sqrt{10} \times \sqrt{2} = \sqrt{10 \times 2} = \sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}$

(2) $\sqrt{21} \div \sqrt{7} = \sqrt{\frac{21}{7}} = \sqrt{3}$

(3) $2\sqrt{6} - 9\sqrt{6} = (2-9)\sqrt{6} = -7\sqrt{6}$

(4) $-\sqrt{27} + \sqrt{75} = -\sqrt{9 \times 3} + \sqrt{25 \times 3} = -3\sqrt{3} + 5\sqrt{3} = (-3+5)\sqrt{3} = 2\sqrt{3}$

(5) $(\sqrt{2}+1)^2 = (\sqrt{2})^2 + 2 \times \sqrt{2} \times 1 + 1^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2}$

3. 最終的な答え

(1) $2\sqrt{5}$

(2) $\sqrt{3}$

(3) $-7\sqrt{6}$

(4) $2\sqrt{3}$

(5) $3 + 2\sqrt{2}$