与えられた4つの式をそれぞれ簡単にせよ。 (1) $\sqrt{24}$ (2) $\sqrt{1700}$ (3) $\sqrt{8} + \sqrt{18} - \sqrt{72}$ (4) $\sqrt{72} - \sqrt{75} + \sqrt{108} - \sqrt{128}$

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) $\sqrt{24}$

(2) $\sqrt{1700}$

(3) $\sqrt{8} + \sqrt{18} - \sqrt{72}$

(4) $\sqrt{72} - \sqrt{75} + \sqrt{108} - \sqrt{128}$

2. 解き方の手順

(1) $\sqrt{24}$

$24 = 4 \times 6 = 2^2 \times 6$ であるから、

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

(2) $\sqrt{1700}$

$1700 = 100 \times 17 = 10^2 \times 17$ であるから、

$\sqrt{1700} = \sqrt{10^2 \times 17} = 10\sqrt{17}$

(3) $\sqrt{8} + \sqrt{18} - \sqrt{72}$

$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$

$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$

$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$

よって、

$\sqrt{8} + \sqrt{18} - \sqrt{72} = 2\sqrt{2} + 3\sqrt{2} - 6\sqrt{2} = (2 + 3 - 6)\sqrt{2} = -1\sqrt{2} = -\sqrt{2}$

(4) $\sqrt{72} - \sqrt{75} + \sqrt{108} - \sqrt{128}$

$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$

$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$

$\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}$

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

よって、

$\sqrt{72} - \sqrt{75} + \sqrt{108} - \sqrt{128} = 6\sqrt{2} - 5\sqrt{3} + 6\sqrt{3} - 8\sqrt{2} = (6 - 8)\sqrt{2} + (-5 + 6)\sqrt{3} = -2\sqrt{2} + \sqrt{3}$

3. 最終的な答え

(1) $2\sqrt{6}$

(2) $10\sqrt{17}$

(3) $-\sqrt{2}$

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