Задание 20, стр 26

Question

Сократите дробь.
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a) $\frac{\sqrt{6}-\sqrt{2}}{\sqrt{8}}$
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b) $\frac{\sqrt{6}+\sqrt{15}}{\sqrt{8}+\sqrt{20}}$
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c) $\frac{10-2 \sqrt{5}}{\sqrt{20}-2}$
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d) $\frac{4+2 \sqrt{2}}{\sqrt{8}+2}$

Gimli · Accepted Answer

$$	ext{a) $\frac{\sqrt{6}-\sqrt{2}}{\sqrt{8}}=\frac{\sqrt{2}(\sqrt{3}-1)}{\sqrt{8}}=\frac{\sqrt{2}}{\sqrt{8}} \cdot(\sqrt{3}-1)=\sqrt{\frac{2}{8}} \cdot(\sqrt{3}-1)=\sqrt{\frac{1}{4}} \cdot(\sqrt{3}-1)=\frac{1}{2}(\sqrt{3}-1)=\frac{\sqrt{3}}{2}-\frac{1}{2}$.}$$
$$	ext{b) $\frac{\sqrt{6}+\sqrt{15}}{\sqrt{8}+\sqrt{20}}=\frac{\sqrt{3}(\sqrt{2}+\sqrt{5})}{\sqrt{4}(\sqrt{2}+\sqrt{5})}=\frac{\sqrt{3}}{\sqrt{4}}=\frac{\sqrt{3}}{2}$.}$$
$$	ext{c) $\frac{10-2 \sqrt{5}}{\sqrt{20}-2}=\frac{2(5-\sqrt{5})}{\sqrt{4}-\sqrt{5}-2}=\frac{2(5-\sqrt{5})}{2 \sqrt{5}-2}=\frac{2 \cdot \sqrt{5}(\sqrt{5}-1)}{2(\sqrt{5}-1)}=\frac{2 \sqrt{5}}{2}=\sqrt{5}$.}$$
$$	ext{d) $\frac{4+2 \sqrt{2}}{\sqrt{8}+2}=\frac{4+2 \sqrt{2}}{\sqrt{4} \cdot \sqrt{2}+2}=\frac{2(2+\sqrt{2})}{2 \sqrt{2}+2}=\frac{(2+\sqrt{2}) 2}{2(\sqrt{2}+1)}=\frac{\sqrt{2}(\sqrt{2}+1) \cdot 2}{2 \cdot(\sqrt{2}+1)}=\sqrt{2}$.}$$

Задание 20, стр 26

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Задание 20, стр 26

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