\[\boxed{\text{160\ (160).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \sqrt[4]{16} = \sqrt[4]{2^{4}} = 2\]
\[\textbf{б)}\ \sqrt[5]{32} = \sqrt[5]{2^{5}} = 2\]
\[\textbf{в)}\ \sqrt[12]{1} = 1\]
\[\textbf{г)}\ \sqrt[3]{- \frac{1}{8}} = \ \sqrt[3]{\left( - \frac{1}{2} \right)^{3}} = - \frac{1}{2}\]
\[\textbf{д)}\ \sqrt[4]{5\frac{1}{16}} = \sqrt[4]{\frac{81}{16}} = \sqrt[4]{\left( \frac{3}{2} \right)^{4}} = \frac{3}{2}\]
\[\textbf{е)}\ \sqrt[3]{3\frac{3}{8}} = \sqrt[3]{\frac{27}{8}} = \sqrt[3]{\left( \frac{3}{2} \right)^{3}} = \frac{3}{2}\ \]
\[\boxed{\text{160.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[Графики\ имеют\ только\ одну\ \]
\[общую\ точку,\ если\ D = 0:\]
\[2x^{2} - 5x + 6 = x^{2} - 7x + n\]
\[x^{2} + 2x + 6 - n = 0\]
\[D = 1 - (6 - n) = 1 - 6 + n =\]
\[= n - 5\]
\[n - 5 = 0\]
\[n = 5.\]
\[x^{2} + 2x + 6 - 5 = 0\]
\[x^{2} + 2x + 1 = 0\]
\[(x + 1)^{2} = 0\]
\[x = - 1\]
\[y( - 1) =\]
\[= 2 \cdot ( - 1)^{2} - 5 \cdot ( - 1) + 6 =\]
\[= 2 + 5 + 6 = 13\]
\[Общая\ точка:\ \ \ ( - 1;13).\]