与えられた4つの式をそれぞれ簡単にせよ。 (1) $\sqrt{12} - \sqrt{27} + 4\sqrt{3}$ (2) $\sqrt{5}\sqrt{30} + 2\sqrt{24} - 3\sqrt{54}$ (3) $(2\sqrt{3} - \sqrt{2})(\sqrt{3} + 3\sqrt{2})$ (4) $(1 + 2\sqrt{5})^2 - (1 - 2\sqrt{5})^2$

1. 問題の内容

与えられた4つの式をそれぞれ簡単にせよ。

(1) $\sqrt{12} - \sqrt{27} + 4\sqrt{3}$

(2) $\sqrt{5}\sqrt{30} + 2\sqrt{24} - 3\sqrt{54}$

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

(4) $(1 + 2\sqrt{5})^2 - (1 - 2\sqrt{5})^2$

2. 解き方の手順

(1)

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$

したがって、

$\sqrt{12} - \sqrt{27} + 4\sqrt{3} = 2\sqrt{3} - 3\sqrt{3} + 4\sqrt{3} = (2 - 3 + 4)\sqrt{3} = 3\sqrt{3}$

(2)

$\sqrt{5}\sqrt{30} = \sqrt{5 \times 30} = \sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6}$

$2\sqrt{24} = 2\sqrt{4 \times 6} = 2 \times 2\sqrt{6} = 4\sqrt{6}$

$3\sqrt{54} = 3\sqrt{9 \times 6} = 3 \times 3\sqrt{6} = 9\sqrt{6}$

したがって、

$\sqrt{5}\sqrt{30} + 2\sqrt{24} - 3\sqrt{54} = 5\sqrt{6} + 4\sqrt{6} - 9\sqrt{6} = (5 + 4 - 9)\sqrt{6} = 0\sqrt{6} = 0$

(3)

$(2\sqrt{3} - \sqrt{2})(\sqrt{3} + 3\sqrt{2}) = 2\sqrt{3} \times \sqrt{3} + 2\sqrt{3} \times 3\sqrt{2} - \sqrt{2} \times \sqrt{3} - \sqrt{2} \times 3\sqrt{2}$

$= 2 \times 3 + 6\sqrt{6} - \sqrt{6} - 3 \times 2 = 6 + 6\sqrt{6} - \sqrt{6} - 6 = 5\sqrt{6}$

(4)

$(1 + 2\sqrt{5})^2 = 1^2 + 2 \times 1 \times 2\sqrt{5} + (2\sqrt{5})^2 = 1 + 4\sqrt{5} + 4 \times 5 = 1 + 4\sqrt{5} + 20 = 21 + 4\sqrt{5}$

$(1 - 2\sqrt{5})^2 = 1^2 - 2 \times 1 \times 2\sqrt{5} + (2\sqrt{5})^2 = 1 - 4\sqrt{5} + 4 \times 5 = 1 - 4\sqrt{5} + 20 = 21 - 4\sqrt{5}$

したがって、

$(1 + 2\sqrt{5})^2 - (1 - 2\sqrt{5})^2 = (21 + 4\sqrt{5}) - (21 - 4\sqrt{5}) = 21 + 4\sqrt{5} - 21 + 4\sqrt{5} = 8\sqrt{5}$

3. 最終的な答え

(1) $3\sqrt{3}$

(2) $0$

(3) $5\sqrt{6}$

(4) $8\sqrt{5}$