Упр.16.83 ГДЗ Мордкович 8 класс (Алгебра)

Задача

а) $$\sqrt{\frac{1}{6}}+\sqrt{\frac{3}{2}}+\sqrt{\frac{2}{3}}-\sqrt{54}$$;

б) $$0{,}1\sqrt{140}-\sqrt{\frac{7}{5}}-\sqrt{\frac{5}{7}}$$;

в) $$\sqrt{18}-\sqrt{\frac{2}{9}}-\sqrt{\frac{9}{2}}$$;

г) $$\sqrt{\frac{1}{14}}+2\sqrt{\frac{2}{7}}-\sqrt{\frac{7}{2}}-\sqrt{14}$$.

Подробный ответ

а)

$$\sqrt{\frac16}+\sqrt{\frac32}+\sqrt{\frac23}-\sqrt{54} = \frac1{\sqrt6}+\frac{\sqrt3}{\sqrt2}+\frac{\sqrt2}{\sqrt3}-3\sqrt6$$
Приведём к общему знаменателю $$\sqrt6$$:
$$\frac1{\sqrt6}+\frac{\sqrt3}{\sqrt2}+\frac{\sqrt2}{\sqrt3}-3\sqrt6 = \frac{1+3+2-18}{\sqrt6} = \frac{-12}{\sqrt6} = -2\sqrt6.$$

б)

$$0{,}1\sqrt{140}-\sqrt{\frac75}-\sqrt{\frac57} = 0{,}1\cdot 2\sqrt{35}-\frac{\sqrt7}{\sqrt5}-\frac{\sqrt5}{\sqrt7}$$
$$= 0{,}2\sqrt{35}-\frac{\sqrt7}{\sqrt5}-\frac{\sqrt5}{\sqrt7} = \frac{0{,}2\cdot 35-7-5}{\sqrt{35}} = \frac{7-12}{\sqrt{35}} = -\frac5{\sqrt{35}} = -\frac{\sqrt{35}}7.$$

в)

$$\sqrt{18}-\sqrt{\frac29}-\sqrt{\frac92} = 3\sqrt2-\frac{\sqrt2}{3}-\frac{3}{\sqrt2}$$
Приведём к общему знаменателю $$3\sqrt2$$:
$$3\sqrt2-\frac{\sqrt2}{3}-\frac{3}{\sqrt2} = \frac{18-2-9}{3\sqrt2} = \frac7{3\sqrt2} = \frac{7\sqrt2}{6}.$$

г)

$$\sqrt{\frac1{14}}+2\sqrt{\frac27}-\sqrt{\frac72}-\sqrt{14} = \frac1{\sqrt{14}}+\frac{2\sqrt2}{\sqrt7}-\frac{\sqrt7}{\sqrt2}-\sqrt{14}$$
Приведём к общему знаменателю $$\sqrt{14}$$:
$$\frac1{\sqrt{14}}+\frac{2\sqrt2}{\sqrt7}-\frac{\sqrt7}{\sqrt2}-\sqrt{14} = \frac{1+4-7-14}{\sqrt{14}} = \frac{-16}{\sqrt{14}} = -\frac{8\sqrt{14}}7.$$