Question

Надо сделать только 2 и 4 !!!!! Очень срочно !!!!! Даю 25 баллов!!!!!

Answer 1

$\displaystyle 2) \\ a).. \frac{5}{\sqrt[3]{4}}= \frac{5*( \sqrt[3]{4})^{2}}{ (\sqrt[3]{4})^{3}}= \frac{5* \sqrt[3]{16}}{4}= \frac{5*2 \sqrt[3]{2} }{4}=2,5 \sqrt[3]{2} \\ \\ b).. \frac{ \sqrt[3]{6}}{ \sqrt[3]{6}+1}= \frac{ \sqrt[3]{6}*( \sqrt[3]{6^{2}}- \sqrt[3]{6}+1) }{ (\sqrt[3]{6}+1)( \sqrt[3]{6^{2}}- \sqrt[3]{6}+1)}= \frac{ \sqrt[3]{6^{3}}- \sqrt[3]{6^{2}}+ \sqrt[3]{6} }{ (\sqrt[3]{6})^{3}+1 }= \\ \\ = \frac{6- \sqrt[3]{36}+ \sqrt[3]{6}}{7}$
$\displaystyle c).. \frac{3}{ \sqrt[3]{49}+ \sqrt[3]{7}+1}= \frac{3*( \sqrt[3]{7}-1)}{( \sqrt[3]{7}-1)(\sqrt[3]{7^{2}}+\sqrt[3]{7}+1)}=\frac{3*(\sqrt[3]{7}-1)}{(\sqrt[3]{7})^{3}-1}=\\\\=\frac{3*(\sqrt[3]{7}-1)}{6}=\frac{\sqrt[3]{7}-1}{2}\\\\\\4..(\sqrt[6]{a}- \sqrt[6]{b})( \sqrt[3]{a}+\sqrt[6]{ab}+\sqrt[3]{b})(\sqrt{a}+\sqrt{b})=\\=((\sqrt[6]{a})^{3}-( \sqrt[6]{b})^{3})( \sqrt{a}+ \sqrt{b})=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})= \\=( \sqrt{a})^{2}-(\sqrt{b})^{2} =a-b$