Решебник по алгебре 9 класс Мерзляк Задание 764

 

\[\boxed{\mathbf{764\ (764).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[b_{1} = 19;\ \ b_{6} = 14\]

\[S_{6} = \frac{b_{1} + b_{6}}{2} \cdot 6 = \frac{19 + 14}{2} \cdot 6 =\]

\[= 33 \cdot 3 = 99.\]

\[\boxed{\mathbf{764.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ x^{2} - 6x + \frac{2}{x - 2} = \frac{2}{x - 2} - 8;\ \ x \neq 2\]

\[x^{2} - 6x + 8 = 0\]

\[D_{1} = 9 - 8 = 1\]

\[x_{1} = 3 + 1 = 4;\]

\[x_{2} = 3 - 1 = 2\ (не\ подходит).\]

\[Ответ:x = 4.\]

\[2)\ \left( \sqrt{x} - 5 \right)\left( 15x^{2} - 7x - 2 \right) = 0;\ \ x \geq 0\]

\[\sqrt{x} - 5 = 0\]

\[\sqrt{x} = 5\]

\[x = 25.\]

\[15x^{2} - 7x - 2 = 0\]

\[D = 49 + 120 = 169\]

\[x_{1} = \frac{7 + 13}{30} = \frac{20}{30} = \frac{2}{3};\]

\[x_{2} = \frac{7 - 13}{30} = - \frac{6}{30} = - \frac{1}{5} < 0.\]

\[Ответ:\ \ x = \frac{2}{3};\ \ x = 25.\]

\[3)\ \left( x^{2} + 6x \right)\left( \sqrt{x} - 4 \right)\left( x^{2} - 8x - 48 \right) = 0\]

\[ОДЗ:\ \ x \geq 0.\]

\[x(x + 6) = 0\]

\[x = 0;\ \ x = - 6 < 0.\]

\[\sqrt{x} - 4 = 0\]

\[\sqrt{x} = 4\]

\[x = 16.\]

\[x^{2} - 8x - 48 = 0\]

\[D_{1} = 16 + 48 = 64\]

\[x_{1} = 4 + 8 = 12;\]

\[x_{2} = 4 - 8 = - 4 < 0.\]

\[Ответ:x = 0;\ \ x = 12;\ \ x = 16.\]