Упр.20 Повторение ГДЗ Колмогоров 10-11 класс (Алгебра)

а) (v3+v2)/(v3-v2)-2v6;
б) (v2+1)^2+(1-v2)^2-(v7+1)(v7-1);
в) (v7+v5)/(v7-v5)-v35;
г) (3v18+2v8+4v50):v2.

а)

$$ \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-2\sqrt{6} = \frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}-2\sqrt{6} $$

$$ = \frac{3+2\sqrt{6}+2}{3-2}-2\sqrt{6} = (5+2\sqrt{6})-2\sqrt{6} = 5 \in \mathbb{Q}. $$

б)

$$ (\sqrt{2}+1)^2+(1-\sqrt{2})^2-(\sqrt{7}+1)(\sqrt{7}-1) $$

$$ = (2+2\sqrt{2}+1)+(1-2\sqrt{2}+2)-(7-1) = 0 \in \mathbb{Q}. $$

в)

$$ \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}-\sqrt{35} = \frac{(\sqrt{7}+\sqrt{5})^2}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}-\sqrt{35} $$

$$ = \frac{7+2\sqrt{35}+5}{7-5}-\sqrt{35} = (6+\sqrt{35})-\sqrt{35} = 6 \in \mathbb{Q}. $$

г)

$$ (3\sqrt{18}+2\sqrt{8}+4\sqrt{50}) : \sqrt{2} = 3\sqrt{9}+2\sqrt{4}+4\sqrt{25} $$

$$ = 3\cdot 3 + 2\cdot 2 + 4\cdot 5 = 9+4+20 = 33 \in \mathbb{Q}. $$

Ответ

а) $$5$$; б) $$0$$; в) $$6$$; г) $$33$$.