問題185の次の式を計算します。 (1) $\sqrt{16} + \sqrt{128}$ (2) $\sqrt{243} + \sqrt{3} - \sqrt{48}$ (3) $\sqrt{192} - \sqrt{81} + \sqrt{\frac{1}{9}}$ (4) $(\sqrt{2} - \sqrt{16})^2$ (5) $\sqrt{16} \times \sqrt{24} \div \sqrt{6}$ (6) $\sqrt[3]{243} \div \sqrt{3} \times \sqrt{9}$

1. 問題の内容

問題185の次の式を計算します。

(1) $\sqrt{16} + \sqrt{128}$

(2) $\sqrt{243} + \sqrt{3} - \sqrt{48}$

(3) $\sqrt{192} - \sqrt{81} + \sqrt{\frac{1}{9}}$

(4) $(\sqrt{2} - \sqrt{16})^2$

(5) $\sqrt{16} \times \sqrt{24} \div \sqrt{6}$

(6) $\sqrt[3]{243} \div \sqrt{3} \times \sqrt{9}$

2. 解き方の手順

(1)

$\sqrt{16} = 4$

$\sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2}$

$\sqrt{16} + \sqrt{128} = 4 + 8\sqrt{2}$

(2)

$\sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3}$

$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$

$\sqrt{243} + \sqrt{3} - \sqrt{48} = 9\sqrt{3} + \sqrt{3} - 4\sqrt{3} = (9+1-4)\sqrt{3} = 6\sqrt{3}$

(3)

$\sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$

$\sqrt{81} = 9$

$\sqrt{\frac{1}{9}} = \frac{1}{3}$

$\sqrt{192} - \sqrt{81} + \sqrt{\frac{1}{9}} = 8\sqrt{3} - 9 + \frac{1}{3} = 8\sqrt{3} - \frac{26}{3}$

(4)

$(\sqrt{2} - \sqrt{16})^2 = (\sqrt{2} - 4)^2 = (\sqrt{2})^2 - 2 \times 4 \times \sqrt{2} + 4^2 = 2 - 8\sqrt{2} + 16 = 18 - 8\sqrt{2}$

(5)

$\sqrt{16} = 4$

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

$\sqrt{16} \times \sqrt{24} \div \sqrt{6} = 4 \times 2\sqrt{6} \div \sqrt{6} = 4 \times 2 = 8$

(6)

$\sqrt[3]{243} = \sqrt[3]{3^5} = \sqrt[3]{3^3 \times 3^2} = 3\sqrt[3]{9}$

$\sqrt{3}$

$\sqrt{9} = 3$

$\sqrt[3]{243} \div \sqrt{3} \times \sqrt{9} = 3\sqrt[3]{9} \div \sqrt{3} \times 3 = 9 \times \frac{\sqrt[3]{9}}{\sqrt{3}} = 9 \times \frac{3^{\frac{2}{3}}}{3^{\frac{1}{2}}} = 9 \times 3^{\frac{2}{3} - \frac{1}{2}} = 9 \times 3^{\frac{4-3}{6}} = 9 \times 3^{\frac{1}{6}} = 9 \sqrt[6]{3}$

3. 最終的な答え

(1) $4 + 8\sqrt{2}$

(2) $6\sqrt{3}$

(3) $8\sqrt{3} - \frac{26}{3}$

(4) $18 - 8\sqrt{2}$

(5) $8$

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