Question

Как это решается, помогите пожалуйста:

Answer 1

$\displaystyle\tt \boxed{5.} \ \ (2^{\frac{5}{3}}\cdot 3^{-\frac{1}{3}}-3^{\frac{5}{3}}\cdot 2^{-\frac{1}{3}})\sqrt[3]{6}=\bigg(\sqrt[3]{2^5}\cdot\frac{1}{\sqrt[3]{3}}-\sqrt[3]{3^5}\cdot \frac{1}{\sqrt[3]{2}}\bigg)\sqrt[3]{6}=\\\\\\ = \bigg(\sqrt[3]{\frac{2^5}{3}}-\sqrt[3]{\frac{3^5}{2}} \bigg)\sqrt[3]{6}=\sqrt[3]{\frac{2^5\cdot6}{3}}-\sqrt[3]{\frac{3^5\cdot6}{2}}=\sqrt[3]{2^5\cdot2}-\sqrt[3]{3^5\cdot3}=\\\\\\=\sqrt[3]{2^6}-\sqrt[3]{3^6}=2^2-3^2=4-9=-5$

$\displaystyle\tt \boxed{6.} \ \ \frac{b}{a-b}+\frac{b^{\frac{1}{2}}}{a^{\frac{1}{2}}+b^{\frac{1}{2}}}=\frac{b}{a-b}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}} =\\\\\\=\frac{b}{a-b}+\frac{\sqrt{b}(\sqrt{a}-\sqrt{b}) }{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}= \frac{b}{a-b}+\frac{\sqrt{ab}-b}{a-b}=\\\\\\=\frac{b+\sqrt{ab}-b }{a-b}=\frac{\sqrt{ab}}{a-b}$