画像にある数学の問題を解く。具体的には、3つの分母の有理化の問題と9つの根号を含む計算問題である。

1. 問題の内容

画像にある数学の問題を解く。具体的には、3つの分母の有理化の問題と9つの根号を含む計算問題である。

2. 解き方の手順

3. (1) 分母を有理化する: $\frac{3\sqrt{5}}{\sqrt{12}}$

$\sqrt{12} = 2\sqrt{3}$なので、

$\frac{3\sqrt{5}}{2\sqrt{3}} = \frac{3\sqrt{5}}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{15}}{2 \times 3} = \frac{\sqrt{15}}{2}$

4. (1) $\sqrt{15} \times \sqrt{80}$

$\sqrt{15} \times \sqrt{80} = \sqrt{15 \times 80} = \sqrt{1200} = \sqrt{400 \times 3} = 20\sqrt{3}$

5. (2) $\sqrt{\frac{75}{2}} \times \sqrt{12}$

$\sqrt{\frac{75}{2}} \times \sqrt{12} = \sqrt{\frac{75}{2} \times 12} = \sqrt{75 \times 6} = \sqrt{25 \times 3 \times 6} = \sqrt{25 \times 18} = 5\sqrt{18} = 5\sqrt{9 \times 2} = 5 \times 3\sqrt{2} = 15\sqrt{2}$

6. (3) $\sqrt{15} \div \frac{\sqrt{3}}{\sqrt{5}}$

$\sqrt{15} \div \frac{\sqrt{3}}{\sqrt{5}} = \sqrt{15} \times \frac{\sqrt{5}}{\sqrt{3}} = \frac{\sqrt{15} \times \sqrt{5}}{\sqrt{3}} = \frac{\sqrt{75}}{\sqrt{3}} = \sqrt{\frac{75}{3}} = \sqrt{25} = 5$

7. (4) $\sqrt{8} \times 2\sqrt{14} \times \sqrt{7}$

$\sqrt{8} \times 2\sqrt{14} \times \sqrt{7} = 2\sqrt{2} \times 2\sqrt{14} \times \sqrt{7} = 4 \times \sqrt{2 \times 14 \times 7} = 4 \times \sqrt{2 \times 2 \times 7 \times 7} = 4 \times 2 \times 7 = 56$

8. (5) $\sqrt{52} \times \sqrt{18} \div \sqrt{39}$

$\sqrt{52} \times \sqrt{18} \div \sqrt{39} = \frac{\sqrt{52} \times \sqrt{18}}{\sqrt{39}} = \frac{\sqrt{52 \times 18}}{\sqrt{39}} = \sqrt{\frac{52 \times 18}{39}} = \sqrt{\frac{4 \times 13 \times 2 \times 9}{3 \times 13}} = \sqrt{\frac{4 \times 2 \times 9}{3}} = \sqrt{4 \times 2 \times 3} = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$

9. (6) $\sqrt{27} \div 2\sqrt{45} \times \sqrt{32}$

$\sqrt{27} \div 2\sqrt{45} \times \sqrt{32} = \frac{\sqrt{27} \times \sqrt{32}}{2\sqrt{45}} = \frac{\sqrt{27 \times 32}}{2\sqrt{45}} = \frac{\sqrt{3^3 \times 2^5}}{2\sqrt{3^2 \times 5}} = \frac{\sqrt{3^2 \times 3 \times 2^4 \times 2}}{2 \times 3 \sqrt{5}} = \frac{3 \times 4 \sqrt{6}}{6 \sqrt{5}} = \frac{12 \sqrt{6}}{6 \sqrt{5}} = \frac{2 \sqrt{6}}{\sqrt{5}} = \frac{2 \sqrt{6} \times \sqrt{5}}{5} = \frac{2 \sqrt{30}}{5}$

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0. (7) $\sqrt{60} \div \sqrt{28} \div \sqrt{15}$

$\sqrt{60} \div \sqrt{28} \div \sqrt{15} = \frac{\sqrt{60}}{\sqrt{28} \times \sqrt{15}} = \frac{\sqrt{60}}{\sqrt{28 \times 15}} = \sqrt{\frac{60}{28 \times 15}} = \sqrt{\frac{60}{420}} = \sqrt{\frac{1}{7}} = \frac{1}{\sqrt{7}} = \frac{\sqrt{7}}{7}$

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1. (8) $\sqrt{135} \div \sqrt{5} \times \sqrt{\frac{98}{3}}$

$\sqrt{135} \div \sqrt{5} \times \sqrt{\frac{98}{3}} = \frac{\sqrt{135}}{\sqrt{5}} \times \sqrt{\frac{98}{3}} = \sqrt{\frac{135}{5}} \times \sqrt{\frac{98}{3}} = \sqrt{27} \times \sqrt{\frac{98}{3}} = \sqrt{27 \times \frac{98}{3}} = \sqrt{9 \times 98} = 3 \sqrt{98} = 3 \sqrt{49 \times 2} = 3 \times 7 \sqrt{2} = 21 \sqrt{2}$

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2. (9) $\frac{2\sqrt{3}}{\sqrt{15}} \div \frac{\sqrt{35}}{5} \times \frac{\sqrt{14}}{4}$

$\frac{2\sqrt{3}}{\sqrt{15}} \div \frac{\sqrt{35}}{5} \times \frac{\sqrt{14}}{4} = \frac{2\sqrt{3}}{\sqrt{15}} \times \frac{5}{\sqrt{35}} \times \frac{\sqrt{14}}{4} = \frac{10\sqrt{3}\sqrt{14}}{4\sqrt{15}\sqrt{35}} = \frac{5\sqrt{3}\sqrt{2}\sqrt{7}}{2\sqrt{3}\sqrt{5}\sqrt{5}\sqrt{7}} = \frac{5\sqrt{2}}{2 \times 5} = \frac{\sqrt{2}}{2}$

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3. 最終的な答え

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4. (1) $\frac{\sqrt{15}}{2}$

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5. (1) $20\sqrt{3}$

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6. (2) $15\sqrt{2}$

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7. (3) $5$

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8. (4) $56$

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9. (5) $2\sqrt{6}$

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0. (6) $\frac{2\sqrt{30}}{5}$

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1. (7) $\frac{\sqrt{7}}{7}$

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2. (8) $21\sqrt{2}$

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3. (9) $\frac{\sqrt{2}}{2}$