画像に写っている複数の平方根の計算問題を解きます。

はい、承知いたしました。画像の問題を解きます。

1. 問題の内容

画像に写っている複数の平方根の計算問題を解きます。

2. 解き方の手順

(1) $2\sqrt{3} + 4\sqrt{3}$

$\sqrt{3}$ で括ると、$(2+4)\sqrt{3} = 6\sqrt{3}$

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

$\sqrt{6}$ で括ると、$(3+1)\sqrt{6} = 4\sqrt{6}$

(3) $8\sqrt{2} - 6\sqrt{2}$

$\sqrt{2}$ で括ると、$(8-6)\sqrt{2} = 2\sqrt{2}$

(4) $5\sqrt{3} - \sqrt{3}$

$\sqrt{3}$ で括ると、$(5-1)\sqrt{3} = 4\sqrt{3}$

(5) $4\sqrt{3} - 5\sqrt{7} + 3\sqrt{7} - 6\sqrt{3}$

$\sqrt{3}$ と $\sqrt{7}$ でそれぞれまとめると、

$(4-6)\sqrt{3} + (-5+3)\sqrt{7} = -2\sqrt{3} - 2\sqrt{7}$

(6) $2\sqrt{5} + \sqrt{3} + 3\sqrt{5} - 5\sqrt{3}$

$\sqrt{5}$ と $\sqrt{3}$ でそれぞれまとめると、

$(2+3)\sqrt{5} + (1-5)\sqrt{3} = 5\sqrt{5} - 4\sqrt{3}$

(7) $\sqrt{12} + \sqrt{75}$

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

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

よって、$2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$

(8) $\sqrt{48} + \sqrt{12}$

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

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

よって、$4\sqrt{3} + 2\sqrt{3} = 6\sqrt{3}$

(9) $\sqrt{27} - \sqrt{12}$

$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

よって、$3\sqrt{3} - 2\sqrt{3} = \sqrt{3}$

(10) $\sqrt{18} - \sqrt{8}$

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

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

よって、$3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$

(11) $\sqrt{28} - \sqrt{54} - \sqrt{24} + \sqrt{63}$

$\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$

$\sqrt{54} = \sqrt{9 \times 6} = 3\sqrt{6}$

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

$\sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7}$

よって、$2\sqrt{7} - 3\sqrt{6} - 2\sqrt{6} + 3\sqrt{7} = 5\sqrt{7} - 5\sqrt{6}$

(12) $\sqrt{72} + \sqrt{24} - \sqrt{18} - \sqrt{96}$

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

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

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

$\sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6}$

よって、$6\sqrt{2} + 2\sqrt{6} - 3\sqrt{2} - 4\sqrt{6} = 3\sqrt{2} - 2\sqrt{6}$

(13) $\sqrt{45} - 4\sqrt{3} - \sqrt{20} + \sqrt{12}$

$\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$

$\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

よって、$3\sqrt{5} - 4\sqrt{3} - 2\sqrt{5} + 2\sqrt{3} = \sqrt{5} - 2\sqrt{3}$

(14) $4\sqrt{12} + \sqrt{20} - \sqrt{75} - 2\sqrt{45}$

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

$\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$

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

$\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$

よって、$4(2\sqrt{3}) + 2\sqrt{5} - 5\sqrt{3} - 2(3\sqrt{5}) = 8\sqrt{3} + 2\sqrt{5} - 5\sqrt{3} - 6\sqrt{5} = 3\sqrt{3} - 4\sqrt{5}$

(15) $5\sqrt{2} + \frac{4}{\sqrt{2}}$

$\frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$

よって、$5\sqrt{2} + 2\sqrt{2} = 7\sqrt{2}$

(16) $\sqrt{45} + \sqrt{5} + \frac{10}{\sqrt{5}}$

$\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$

$\frac{10}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}$

よって、$3\sqrt{5} + \sqrt{5} + 2\sqrt{5} = 6\sqrt{5}$

(17) $-\frac{3}{\sqrt{2}} + \frac{\sqrt{50}}{4}$

$-\frac{3}{\sqrt{2}} = -\frac{3\sqrt{2}}{2}$

$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$

$\frac{\sqrt{50}}{4} = \frac{5\sqrt{2}}{4}$

よって,$-\frac{3\sqrt{2}}{2} + \frac{5\sqrt{2}}{4} = -\frac{6\sqrt{2}}{4} + \frac{5\sqrt{2}}{4} = -\frac{\sqrt{2}}{4}$

(18) $\frac{5}{\sqrt{6}} - \frac{\sqrt{24}}{4}$

$\frac{5}{\sqrt{6}} = \frac{5\sqrt{6}}{6}$

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

$\frac{\sqrt{24}}{4} = \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2} = \frac{3\sqrt{6}}{6}$

よって、$\frac{5\sqrt{6}}{6} - \frac{3\sqrt{6}}{6} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3}$

3. 最終的な答え

(1) $6\sqrt{3}$

(2) $4\sqrt{6}$

(3) $2\sqrt{2}$

(4) $4\sqrt{3}$

(5) $-2\sqrt{3} - 2\sqrt{7}$

(6) $5\sqrt{5} - 4\sqrt{3}$

(7) $7\sqrt{3}$

(8) $6\sqrt{3}$

(9) $\sqrt{3}$

(10) $\sqrt{2}$

(11) $5\sqrt{7} - 5\sqrt{6}$

(12) $3\sqrt{2} - 2\sqrt{6}$

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

(14) $3\sqrt{3} - 4\sqrt{5}$

(15) $7\sqrt{2}$

(16) $6\sqrt{5}$

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

(18) $\frac{\sqrt{6}}{3}$