Задание 17, стр 26

Question

Упростите .
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a) $(\sqrt{6}+\sqrt{5})^{2}-\sqrt{120}$
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b) $\sqrt{60}+(\sqrt{5}-\sqrt{3})^{2}$
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c) $(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})^{2}$
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d) $(\sqrt{3+2 \sqrt{2}}-\sqrt{3-2 \sqrt{2}})^{2}$

Gimli · Accepted Answer

a) $$(\sqrt{6}+\sqrt{5})^{2}-\sqrt{120}=(\sqrt{6})^{2}+2 \cdot \sqrt{6} \cdot \sqrt{5}+(\sqrt{5})^{2}-\sqrt{120}=6+2 \sqrt{30}+5-\sqrt{120}=6+\sqrt{4 \cdot 30}+5-$$
$$-\sqrt{120}\cdot 6+\sqrt{120}+5-\sqrt{120}=11.$$

b) $$\sqrt{60}+(\sqrt{5}-\sqrt{3})^{2}=\sqrt{60}+(\sqrt{5})^{2}-2 \sqrt{5} \cdot \sqrt{3}+(\sqrt{3})^2=\sqrt{60}+5-2 \sqrt{15}+3=\sqrt{60}+5-\sqrt{4 \cdot 15}+3=$$
$$=\sqrt{60}+5-\sqrt{60}+3=8.$$

c) $$(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})^{2}=(\sqrt{2+\sqrt{3}})^{2}+2 \cdot \sqrt{(2+\sqrt{3})(2-\sqrt{3})}+(\sqrt{2-\sqrt{3}})^{2}=2+\sqrt{3}+2 \sqrt{4-3}+2-$$
$$-\sqrt{3}=2+\sqrt{3}+2+2-\sqrt{3}=2+2+2=6$$

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

Задание 17, стр 26

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Задание 17, стр 26

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