Question

Освободить от иррациональности знаменатилей.
4/√7+3
10/√10-√2

Answer 1

Ответ:

$6-2\sqrt{7}; \quad \frac{5 \cdot (\sqrt{10}+\sqrt{2})}{4};$

Объяснение:

$\frac{4}{\sqrt{7}+3}=\frac{4 \cdot (\sqrt{7}-3)}{(\sqrt{7}+3) \cdot (\sqrt{7}-3)}=\frac{4 \cdot (\sqrt{7}-3)}{(\sqrt{7})^{2}-3^{2}}=\frac{4 \cdot (\sqrt{7}-3)}{7-9}=\frac{4 \cdot (\sqrt{7}-3)}{-2}=-2 \cdot (\sqrt{7}-3)=$

$=2 \cdot (3-\sqrt{7})=6-2\sqrt{7};$

$\frac{10}{\sqrt{10}-\sqrt{2}}=\frac{10 \cdot (\sqrt{10}+\sqrt{2})}{(\sqrt{10}-\sqrt{2})\cdot (\sqrt{10}+\sqrt{2})}=\frac{10 \cdot (\sqrt{10}+\sqrt{2})}{(\sqrt{10})^{2}-(\sqrt{2})^{2}}=\frac{10 \cdot (\sqrt{10}+\sqrt{2})}{10-2}=\frac{10 \cdot (\sqrt{10}+\sqrt{2})}{8}=$

$=\frac{5 \cdot (\sqrt{10}+\sqrt{2})}{4};$