Question

Прошу, помогите с заданием.

Answer 1

1
$\frac{1}{ \sqrt[3]{9} } = \frac{ \sqrt[3]{3} }{ \sqrt[ 3]{9} \times \sqrt[3]{3} } = \frac{ \sqrt[3]{3} }{ \sqrt[3]{27} } = \frac{ \sqrt[3]{3} }{3}$
2.
$\frac{4}{ \sqrt[3]{7} - \sqrt[3]{3} } = \frac{4( \sqrt[3]{ {7}^{2} } + \sqrt[3]{7} \times \sqrt[3]{3} + \sqrt[3]{ {3}^{2}} )}{( \sqrt[3]{7} - \sqrt[3]{3} )(\sqrt[3]{ {7}^{2} } + \sqrt[3]{7} \times \sqrt[3]{3} + \sqrt[3]{ {3}^{2}} )}= \frac{4(\sqrt[3]{ {7}^{2} } + \sqrt[3]{7} \times \sqrt[3]{3} + \sqrt[3]{ {3}^{2}})}{ \sqrt[3]{ {7}^{3} } - \sqrt[3]{ {3}^{3} } } = \frac{4 \times (\sqrt[3]{ {7}^{2} } + \sqrt[3]{7} \times \sqrt[3]{3} + \sqrt[3]{ {3}^{2}})}{7 - 3} = \frac{4(\sqrt[3]{ {7}^{2} } + \sqrt[3]{7} \times \sqrt[3]{3} + \sqrt[3]{ {3}^{2}})}{4} = \sqrt[3]{ {7}^{2} } + \sqrt[3]{7} \times \sqrt[3]{3} + \sqrt[3]{ {3}^{2}} = \sqrt[3]{49} + \sqrt[3]{21} + \sqrt[3]{9}$

Answer 2

$1)\; \; \frac{1}{\sqrt[3]9}=\frac{\sqrt[3]3}{\sqrt[3]{9\cdot 3}} =\frac{\sqrt[3]3}{3}\\\\\\2)\; \; \frac{4}{\sqrt[3]7-\sqrt[3]3}=\frac{4\cdot ((\sqrt[3]7)^2+\sqrt[3]7\vfot \sqrt[3]3+(\sqrt[3]3)^2)}{(\sqrt[3]7)^3-(\sqrt[3]3)^3}=\frac{4\cdot (\sqrt[3]{49}+\sqrt[3]{21}+\sqrt[3]9)}{4}=\\\\=\sqrt[3]{49}+\sqrt[3]{21}+\sqrt[3]9$