\[\boxed{\text{168\ (168).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\mathbf{При\ любом\ нечетном\ }\mathbf{\text{n\ }}\mathbf{и\ }\]
\[\mathbf{четном\ }\mathbf{\text{a\ }}\mathbf{верно\ равенство:}\]
\[\sqrt[\mathbf{n}]{\mathbf{- a}}\mathbf{= -}\sqrt[\mathbf{n}]{\mathbf{a}}\mathbf{.}\]
Решение.
\[\textbf{а)}\ \sqrt[5]{- 32} = - \sqrt[5]{2^{5}} = - 2;\]
\[\textbf{б)}\ \sqrt[7]{- 1} = - 1;\]
\[\textbf{в)} - 2 \cdot \sqrt[4]{81} = - 2 \cdot \sqrt[4]{3^{4}} =\]
\[= - 2 \cdot 3 = - 6;\]
\[\textbf{г)} - 4 \cdot \ \sqrt[3]{27} = - 4 \cdot \sqrt[3]{3^{3}} =\]
\[= - 4 \cdot 3 = - 12;\]
\[\textbf{д)}\ \sqrt[5]{32} + \sqrt[3]{- 8} =\]
\[= \sqrt[5]{2^{5}} + \sqrt[3]{( - 2)^{3}} = 2 - 2 = 0;\]
\[\textbf{е)}\ \sqrt[4]{625} - \sqrt[3]{- 125} =\]
\[= \sqrt[4]{5^{4}} - \sqrt[3]{( - 5)^{3}} = 5 + 5 = 10.\]
\[\boxed{\text{168.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ y = \frac{10}{x - 3} - 2;\]
\[x = 3\ \ \ и\ \ \ y = - 2 \Longrightarrow\]
\[\Longrightarrow асимптоты;\]
\[\textbf{б)}\ y = \frac{8}{x + 2} - 3;\ \]
\[x = - 2\ \ \ и\ \ \ y = - 3 \Longrightarrow\]
\[\Longrightarrow асимптоты.\]