Question

Помогите пж номер 4 Б)

Answer 1

Ответ:

$2\sqrt{6}-2\sqrt{2}$

Объяснение:

$\frac{8}{\sqrt{6}+\sqrt{2}} = \frac{8}{\sqrt{6}+\sqrt{2}} * \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}} = \frac{8(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2}) * (\sqrt{6}-\sqrt{2})} = \frac{8(\sqrt{6}-\sqrt{2})}{\sqrt{6} ^{2}-\sqrt{2} ^{2} } = \frac{8(\sqrt{6}-\sqrt{2})}{6-2} = \frac{8(\sqrt{6}-\sqrt{2})}{4} = 2(\sqrt{6}-\sqrt{2}) = 2\sqrt{6} -2\sqrt{2}$

использовали формулу сокращённого умножения: $(a+b)*(a-b) = a^{2} - b^{2}$