次の計算をせよ。 (1) $\sqrt{10} \times \sqrt{35}$ (2) $\sqrt{20} - \sqrt{45}$ (3) $\sqrt{\frac{27}{16}} + \sqrt{3}$ (4) $\sqrt{21} \times (\sqrt{14} + \sqrt{6})$

1. 問題の内容

次の計算をせよ。

(1) $\sqrt{10} \times \sqrt{35}$

(2) $\sqrt{20} - \sqrt{45}$

(3) $\sqrt{\frac{27}{16}} + \sqrt{3}$

(4) $\sqrt{21} \times (\sqrt{14} + \sqrt{6})$

2. 解き方の手順

(1)

$\sqrt{10} \times \sqrt{35} = \sqrt{10 \times 35} = \sqrt{2 \times 5 \times 5 \times 7} = \sqrt{5^2 \times 2 \times 7} = 5\sqrt{14}$

(2)

$\sqrt{20} - \sqrt{45} = \sqrt{4 \times 5} - \sqrt{9 \times 5} = 2\sqrt{5} - 3\sqrt{5} = -1\sqrt{5} = -\sqrt{5}$

(3)

$\sqrt{\frac{27}{16}} + \sqrt{3} = \frac{\sqrt{27}}{\sqrt{16}} + \sqrt{3} = \frac{\sqrt{9 \times 3}}{4} + \sqrt{3} = \frac{3\sqrt{3}}{4} + \sqrt{3} = \frac{3\sqrt{3}}{4} + \frac{4\sqrt{3}}{4} = \frac{7\sqrt{3}}{4}$

(4)

$\sqrt{21} \times (\sqrt{14} + \sqrt{6}) = \sqrt{21} \times \sqrt{14} + \sqrt{21} \times \sqrt{6} = \sqrt{3 \times 7 \times 2 \times 7} + \sqrt{3 \times 7 \times 2 \times 3} = \sqrt{7^2 \times 3 \times 2} + \sqrt{3^2 \times 7 \times 2} = 7\sqrt{6} + 3\sqrt{14}$

3. 最終的な答え

(1) $5\sqrt{14}$

(2) $-\sqrt{5}$

(3) $\frac{7\sqrt{3}}{4}$

(4) $7\sqrt{6} + 3\sqrt{14}$