Задание 26, стр 27

Question

Докажите равенство.
$$
\frac{1}{3+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{5}}+\frac{1}{\sqrt{5}+2}=1
$$

Gimli · Accepted Answer

$$
\frac{1}{3+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{5}}+\frac{1}{\sqrt{5}+2}=$$
$$=\frac{1 \cdot(3-\sqrt{8})}{(3+\sqrt{8})(3-\sqrt{8})}+\frac{1 \cdot(\sqrt{8}-\sqrt{7})}{(\sqrt{8}+\sqrt{7})(\sqrt{8}-\sqrt{7})}+\frac{1 \cdot(\sqrt{7}-\sqrt{6})}{(\sqrt{7}+\sqrt{6})(\sqrt{7}-\sqrt{6})}+\frac{1 \cdot(\sqrt{6}-\sqrt{5})}{(\sqrt{6}+\sqrt{5})(\sqrt{6}-\sqrt{3})}+\frac{1 \cdot(\sqrt{5}-2)}{(\sqrt{5}+2)(\sqrt{5}-2)}=$$
$$=\frac{3-\sqrt{8}}{9-8}+\frac{\sqrt{8}-\sqrt{7}}{8-7}+\frac{\sqrt{7}-\sqrt{6}}{7-6}+\frac{\sqrt{6}-\sqrt{5}}{6-5}+\frac{\sqrt{5}-\sqrt{2}}{5-4}=$$
$$=3-\sqrt{8}+\sqrt{8}-\sqrt{7}+\sqrt{7}-\sqrt{6}+\sqrt{6}-\sqrt{5}+\sqrt{5} -2=3-2=1.$$
	extbf{Ответ: 1.}

Задание 26, стр 27

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Задание 26, стр 27

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