Упр.2.149 ГДЗ Дорофеев Суворова 8 класс (Алгебра)

Докажите, что:

а) $$\sqrt{3}\cdot\sqrt{3+\sqrt{6}}\cdot\sqrt{3-\sqrt{6}}=3$$;

б) $$\sqrt{2}\cdot\sqrt{2+\sqrt{2}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}=2$$;

в) $$\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=1$$.

а)

$$\sqrt{3}\cdot \sqrt{3+\sqrt{6}}\cdot \sqrt{3-\sqrt{6}} = \sqrt{3(3+\sqrt{6})(3-\sqrt{6})}$$
$$= \sqrt{3(9-6)}=\sqrt{3\cdot 3}=\sqrt{9}=3.$$

б)

$$\sqrt{2}\cdot \sqrt{2+\sqrt{2}}\cdot \sqrt{2+\sqrt{2+\sqrt{2}}}\cdot \sqrt{2-\sqrt{2+\sqrt{2}}}$$
$$= \sqrt{2}\cdot \sqrt{2+\sqrt{2}}\cdot \sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)}$$
$$= \sqrt{2}\cdot \sqrt{2+\sqrt{2}}\cdot \sqrt{4-\left(2+\sqrt{2}\right)}$$
$$= \sqrt{2}\cdot \sqrt{2+\sqrt{2}}\cdot \sqrt{2-\sqrt{2}}$$
$$= \sqrt{2}\cdot \sqrt{(2+\sqrt{2})(2-\sqrt{2})} = \sqrt{2}\cdot \sqrt{4-2}$$
$$= \sqrt{2}\cdot \sqrt{2}=2.$$

в)

$$\sqrt{2+\sqrt{3}}\cdot \sqrt{2+\sqrt{2+\sqrt{3}}}\cdot \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$$
$$= \sqrt{2+\sqrt{3}}\cdot \sqrt{2+\sqrt{2+\sqrt{3}}}\cdot \sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}$$
$$= \sqrt{2+\sqrt{3}}\cdot \sqrt{2+\sqrt{2+\sqrt{3}}}\cdot \sqrt{4-\left(2+\sqrt{2+\sqrt{3}}\right)}$$
$$= \sqrt{2+\sqrt{3}}\cdot \sqrt{2+\sqrt{2+\sqrt{3}}}\cdot \sqrt{2-\sqrt{2+\sqrt{3}}}$$
$$= \sqrt{2+\sqrt{3}}\cdot \sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}$$
$$= \sqrt{2+\sqrt{3}}\cdot \sqrt{4-\left(2+\sqrt{3}\right)} = \sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}$$
$$= \sqrt{(2+\sqrt{3})(2-\sqrt{3})} = \sqrt{4-3} = \sqrt{1}=1.$$

Ответ

а) $$3$$; б) $$2$$; в) $$1$$.