## 1. 問題の内容

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1. 問題の内容

与えられた6つの式を計算する問題です。各式は累乗根を含む加減乗除の計算です。

(1) $\sqrt[3]{18} \times \sqrt[3]{12}$

(2) $\sqrt[4]{27} \times \sqrt{27} \div \sqrt[4]{3}$

(3) $\sqrt[3]{6} \times \sqrt[6]{6} \times \sqrt[4]{12}$

(4) $2\sqrt[4]{5} + 3\sqrt[4]{5}$

(5) $\sqrt[3]{24} + \sqrt[3]{81} - \sqrt[3]{3}$

(6) $\sqrt[3]{54} + \sqrt[3]{16} - \sqrt[3]{\frac{1}{4}}$

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2. 解き方の手順

(1)

$\sqrt[3]{18} \times \sqrt[3]{12} = \sqrt[3]{18 \times 12} = \sqrt[3]{216} = 6$

(2)

$\sqrt[4]{27} \times \sqrt{27} \div \sqrt[4]{3} = \sqrt[4]{\frac{27}{3}} \times \sqrt{27} = \sqrt[4]{9} \times \sqrt{27} = \sqrt{3} \times 3\sqrt{3} = 3 \times 3 = 9$

(3)

$\sqrt[3]{6} \times \sqrt[6]{6} \times \sqrt[4]{12} = 6^{\frac{1}{3}} \times 6^{\frac{1}{6}} \times (2^2 \times 3)^{\frac{1}{4}} = 6^{\frac{1}{3} + \frac{1}{6}} \times 2^{\frac{2}{4}} \times 3^{\frac{1}{4}} = 6^{\frac{1}{2}} \times 2^{\frac{1}{2}} \times 3^{\frac{1}{4}} = \sqrt{6} \times \sqrt{2} \times \sqrt[4]{3} = \sqrt{12} \times \sqrt[4]{3} = 2\sqrt{3} \times \sqrt[4]{3} = 2 \times 3^{\frac{1}{2}} \times 3^{\frac{1}{4}} = 2 \times 3^{\frac{3}{4}} = 2 \sqrt[4]{27}$

(4)

$2\sqrt[4]{5} + 3\sqrt[4]{5} = (2+3)\sqrt[4]{5} = 5\sqrt[4]{5}$

(5)

$\sqrt[3]{24} + \sqrt[3]{81} - \sqrt[3]{3} = \sqrt[3]{8 \times 3} + \sqrt[3]{27 \times 3} - \sqrt[3]{3} = 2\sqrt[3]{3} + 3\sqrt[3]{3} - \sqrt[3]{3} = (2+3-1)\sqrt[3]{3} = 4\sqrt[3]{3}$

(6)

$\sqrt[3]{54} + \sqrt[3]{16} - \sqrt[3]{\frac{1}{4}} = \sqrt[3]{27 \times 2} + \sqrt[3]{8 \times 2} - \sqrt[3]{\frac{1}{4}} = 3\sqrt[3]{2} + 2\sqrt[3]{2} - \sqrt[3]{\frac{1}{4}} = 5\sqrt[3]{2} - \sqrt[3]{\frac{1}{4}} = 5\sqrt[3]{2} - \sqrt[3]{\frac{2}{8}} = 5\sqrt[3]{2} - \frac{1}{2}\sqrt[3]{2} = \frac{9}{2}\sqrt[3]{2}$

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3. 最終的な答え

(1) 6

(2) 9

(3) $2\sqrt[4]{27}$

(4) $5\sqrt[4]{5}$

(5) $4\sqrt[3]{3}$

(6) $\frac{9}{2}\sqrt[3]{2}$