Question

Избавьтесь от иррациональности в знаменателе а) б) в) г) ​

Answer 1

Пошаговое объяснение:

$\displaystyle \frac{6}{ \sqrt{3} } = \frac{6 \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } = \frac{6 \sqrt{3} }{3} = 2 \sqrt{3}$

$\displaystyle \frac{2}{ \sqrt[3]{2} } = \frac{2 \sqrt[3]{4} }{ \sqrt[3]{2} \times \sqrt[3]{2} \times \sqrt[3]{2} } = \frac{2 \sqrt[3]{4} }{2} = \sqrt[3]{4}$

$\displaystyle \frac{3ab}{ \sqrt{3a} } = \frac{3ab \sqrt{3a} }{ \sqrt{3a} \times \sqrt{3a} } = \frac{3ab \sqrt{3a}}{3a} = b \sqrt{3a}$

$\displaystyle \frac{x - y}{ \sqrt{x - y} } = \frac{(x - y) \sqrt{x - y} }{ (\sqrt{x - y} ) \times (\sqrt{x - y} ) } = \frac{(x - y) \sqrt{x - y}}{x - y} = \sqrt{x - y}$

Answer 2

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