与えられた6つの根号を含む計算問題を解く。 (1) $\sqrt{27} \times \sqrt{8}$ (2) $\sqrt{14} \times \sqrt{21}$ (3) $\sqrt{18} \times \sqrt{50}$ (4) $\sqrt{56} \times \sqrt{40}$ (5) $\sqrt{45} \times \sqrt{35}$ (6) $3\sqrt{6} \times 4\sqrt{10}$

1. 問題の内容

与えられた6つの根号を含む計算問題を解く。

(1) $\sqrt{27} \times \sqrt{8}$

(2) $\sqrt{14} \times \sqrt{21}$

(3) $\sqrt{18} \times \sqrt{50}$

(4) $\sqrt{56} \times \sqrt{40}$

(5) $\sqrt{45} \times \sqrt{35}$

(6) $3\sqrt{6} \times 4\sqrt{10}$

2. 解き方の手順

(1) $\sqrt{27} \times \sqrt{8} = \sqrt{3^3} \times \sqrt{2^3} = 3\sqrt{3} \times 2\sqrt{2} = 6\sqrt{6}$

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

(3) $\sqrt{18} \times \sqrt{50} = \sqrt{2 \times 3^2} \times \sqrt{2 \times 5^2} = 3\sqrt{2} \times 5\sqrt{2} = 15 \times 2 = 30$

(4) $\sqrt{56} \times \sqrt{40} = \sqrt{2^3 \times 7} \times \sqrt{2^3 \times 5} = \sqrt{2^6 \times 5 \times 7} = 2^3 \sqrt{35} \times \sqrt{2} = 8\sqrt{35} \times \sqrt{2}= \sqrt{2240} = \sqrt{16 \times 140} = 4 \sqrt{140} = 4 \sqrt{4 \times 35} = 8 \sqrt{35} = \sqrt{56 \times 40} = \sqrt{2240} = \sqrt{16 * 140} = 4 \sqrt{140} = 4 \sqrt{4*35} = 8 \sqrt{35}$

$\sqrt{56}\times \sqrt{40}=\sqrt{4 \times 14} \times \sqrt{4 \times 10} = 2\sqrt{14} \times 2\sqrt{10}= 4\sqrt{140} = 4\sqrt{4 \times 35}= 4 \times 2 \sqrt{35}= 8\sqrt{35}$

(5) $\sqrt{45} \times \sqrt{35} = \sqrt{3^2 \times 5} \times \sqrt{5 \times 7} = \sqrt{3^2 \times 5^2 \times 7} = 3 \times 5 \sqrt{7} = 15\sqrt{7}$

(6) $3\sqrt{6} \times 4\sqrt{10} = 12\sqrt{60} = 12\sqrt{4 \times 15} = 12 \times 2 \sqrt{15} = 24\sqrt{15}$

3. 最終的な答え

(1) $6\sqrt{6}$

(2) $7\sqrt{6}$

(3) $30$

(4) $8\sqrt{35}$

(5) $15\sqrt{7}$

(6) $24\sqrt{15}$