与えられた根号を含む式の計算問題を解きます。具体的には、以下の12個の問題を解きます。 (1) $\sqrt{8} + \sqrt{18}$ (2) $\sqrt{40} + 5\sqrt{10}$ (3) $\sqrt{50} + \sqrt{2/8}$ (4) $\sqrt{8} - \sqrt{32}$ (5) $\sqrt{12} - \sqrt{48}$ (6) $2\sqrt{27} - 4\sqrt{12}$ (7) $\sqrt{32} + \sqrt{8} + \sqrt{18}$ (8) $\sqrt{48} + \sqrt{3} - \sqrt{27}$ (9) $\sqrt{20} + \sqrt{80} - \sqrt{125}$ (10) $\sqrt{50} - \sqrt{12} + \sqrt{27}$ (11) $\sqrt{90} - \sqrt{160} + 6\sqrt{10}$ (12) $3\sqrt{32} - 2\sqrt{18} - \sqrt{8}$

1. 問題の内容

与えられた根号を含む式の計算問題を解きます。具体的には、以下の12個の問題を解きます。

(1) $\sqrt{8} + \sqrt{18}$

(2) $\sqrt{40} + 5\sqrt{10}$

(3) $\sqrt{50} + \sqrt{2/8}$

(4) $\sqrt{8} - \sqrt{32}$

(5) $\sqrt{12} - \sqrt{48}$

(6) $2\sqrt{27} - 4\sqrt{12}$

(7) $\sqrt{32} + \sqrt{8} + \sqrt{18}$

(8) $\sqrt{48} + \sqrt{3} - \sqrt{27}$

(9) $\sqrt{20} + \sqrt{80} - \sqrt{125}$

(10) $\sqrt{50} - \sqrt{12} + \sqrt{27}$

(11) $\sqrt{90} - \sqrt{160} + 6\sqrt{10}$

(12) $3\sqrt{32} - 2\sqrt{18} - \sqrt{8}$

2. 解き方の手順

各根号の中身を素因数分解し、$\sqrt{a^2} = a$ を利用して根号の外に出せる数字を出します。

次に、根号の中身が同じ項同士をまとめます。

(1) $\sqrt{8} + \sqrt{18} = \sqrt{2^2 \cdot 2} + \sqrt{3^2 \cdot 2} = 2\sqrt{2} + 3\sqrt{2} = (2+3)\sqrt{2} = 5\sqrt{2}$

(2) $\sqrt{40} + 5\sqrt{10} = \sqrt{2^2 \cdot 10} + 5\sqrt{10} = 2\sqrt{10} + 5\sqrt{10} = (2+5)\sqrt{10} = 7\sqrt{10}$

(3) $\sqrt{50} + \sqrt{\frac{2}{8}} = \sqrt{5^2 \cdot 2} + \sqrt{\frac{1}{4}} = 5\sqrt{2} + \frac{1}{2} = 5\sqrt{2} + \frac{1}{2}$

(4) $\sqrt{8} - \sqrt{32} = \sqrt{2^2 \cdot 2} - \sqrt{4^2 \cdot 2} = 2\sqrt{2} - 4\sqrt{2} = (2-4)\sqrt{2} = -2\sqrt{2}$

(5) $\sqrt{12} - \sqrt{48} = \sqrt{2^2 \cdot 3} - \sqrt{4^2 \cdot 3} = 2\sqrt{3} - 4\sqrt{3} = (2-4)\sqrt{3} = -2\sqrt{3}$

(6) $2\sqrt{27} - 4\sqrt{12} = 2\sqrt{3^2 \cdot 3} - 4\sqrt{2^2 \cdot 3} = 2 \cdot 3\sqrt{3} - 4 \cdot 2\sqrt{3} = 6\sqrt{3} - 8\sqrt{3} = (6-8)\sqrt{3} = -2\sqrt{3}$

(7) $\sqrt{32} + \sqrt{8} + \sqrt{18} = \sqrt{4^2 \cdot 2} + \sqrt{2^2 \cdot 2} + \sqrt{3^2 \cdot 2} = 4\sqrt{2} + 2\sqrt{2} + 3\sqrt{2} = (4+2+3)\sqrt{2} = 9\sqrt{2}$

(8) $\sqrt{48} + \sqrt{3} - \sqrt{27} = \sqrt{4^2 \cdot 3} + \sqrt{3} - \sqrt{3^2 \cdot 3} = 4\sqrt{3} + \sqrt{3} - 3\sqrt{3} = (4+1-3)\sqrt{3} = 2\sqrt{3}$

(9) $\sqrt{20} + \sqrt{80} - \sqrt{125} = \sqrt{2^2 \cdot 5} + \sqrt{4^2 \cdot 5} - \sqrt{5^2 \cdot 5} = 2\sqrt{5} + 4\sqrt{5} - 5\sqrt{5} = (2+4-5)\sqrt{5} = \sqrt{5}$

(10) $\sqrt{50} - \sqrt{12} + \sqrt{27} = \sqrt{5^2 \cdot 2} - \sqrt{2^2 \cdot 3} + \sqrt{3^2 \cdot 3} = 5\sqrt{2} - 2\sqrt{3} + 3\sqrt{3} = 5\sqrt{2} + (3-2)\sqrt{3} = 5\sqrt{2} + \sqrt{3}$

(11) $\sqrt{90} - \sqrt{160} + 6\sqrt{10} = \sqrt{3^2 \cdot 10} - \sqrt{4^2 \cdot 10} + 6\sqrt{10} = 3\sqrt{10} - 4\sqrt{10} + 6\sqrt{10} = (3-4+6)\sqrt{10} = 5\sqrt{10}$

(12) $3\sqrt{32} - 2\sqrt{18} - \sqrt{8} = 3\sqrt{4^2 \cdot 2} - 2\sqrt{3^2 \cdot 2} - \sqrt{2^2 \cdot 2} = 3 \cdot 4\sqrt{2} - 2 \cdot 3\sqrt{2} - 2\sqrt{2} = 12\sqrt{2} - 6\sqrt{2} - 2\sqrt{2} = (12-6-2)\sqrt{2} = 4\sqrt{2}$

3. 最終的な答え

(1) $5\sqrt{2}$

(2) $7\sqrt{10}$

(3) $5\sqrt{2} + \frac{1}{2}$

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

(5) $-2\sqrt{3}$

(6) $-2\sqrt{3}$

(7) $9\sqrt{2}$

(8) $2\sqrt{3}$

(9) $\sqrt{5}$

(10) $5\sqrt{2} + \sqrt{3}$

(11) $5\sqrt{10}$

(12) $4\sqrt{2}$