Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по алгебре 9 класс Мерзляк Задание 766
Задание 766
\[\boxed{\mathbf{766\ (766).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[- 8;\ - 6;\ - 4;\ldots\]
\[d = - 6 - ( - 8) = - 6 + 8 = 2\]
\[S_{20} = \frac{2a_{1} + 19d}{2} \cdot 20 =\]
\[= \frac{- 16 + 38}{2} \cdot 20 =\]
\[= 11 \cdot 20 = 220.\]
\[\boxed{\mathbf{766.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[- 4x^{2} - 16x + 128 = 0\ \ \ \ |\ :( - 4)\]
\[x^{2} + 4x - 32 = 0\]
\[D_{1} = 4 + 32 = 36\]
\[x_{1} = - 2 + 6 = 4;\]
\[x_{2} = - 2 - 6 = - 8.\]
\[Ответ:\ - 8;\ \ 4.\]
\[2)\ \frac{x^{2} - 3x + 6}{x} + \frac{2x}{x^{2} - 3x + 6} = 3\]
\[y = x^{2} - 3x + 6:\]
\[\frac{y}{x} + \frac{2x}{y} = 3\ \ \ \ | \cdot xy\]
\[y^{2} + 2x^{2} = 3xy;\ \ \ x \neq 0;\ \ y \neq 0.\]
\[Подставим:\]
\[\left( x^{2} - 3x + 6 \right)^{2} + 2x^{2} = 3x\left( x^{2} - 3x + 6 \right)\]
\[x^{4} - 9x^{3} + 32x^{2} - 54x + 36 = 0\]
\[x = 2:\]
\[(x - 2)\left( x^{3} - 7x^{2} + 18x - 18 \right) = 0\]
\[x = 3:\]
\[(x - 2)(x - 3)\left( x^{2} - 4x + 6 \right) = 0\]
\[x^{2} - 4x + 6 = 0\]
\[D_{1} = 4 - 6 < 0\]
\[Нет\ корней.\]
\[Ответ:x = 2;\ \ x = 3.\]
\[3)\ \frac{x^{2}}{(3x - 1)^{2}} - \frac{4x^{\backslash 3x - 1}}{3x - 1} - 5^{{\backslash(3x - 1)}^{2}} = 0\]
\[ОДЗ:\ \ 3x - 1 \neq 0\]
\[3x \neq 1\]
\[x \neq \frac{1}{3}.\]
\[x^{2} - 12x^{2} + 4x - 5\left( 9x^{2} - 6x + 1 \right) = 0\]
\[- 11x^{2} + 4x - 45x^{2} + 30x - 5 = 0\]
\[- 56x^{2} + 34x - 5 = 0\]
\[56x^{2} - 34x + 5 = 0\]
\[D_{1} = 289 - 280 = 9\]
\[x_{1} = \frac{17 + 3}{56} = \frac{20}{56} = \frac{5}{14};\]
\[x_{2} = \frac{17 - 3}{56} = \frac{14}{56} = \frac{1}{4}.\]
\[Ответ:\ \ x = \frac{5}{14};\ \ x = \frac{1}{4}.\]
\[4)\ \frac{24^{\backslash x^{2} + 2x - 3}}{x^{2} + 2x - 8} - \frac{15^{\backslash x^{2} + 2x - 8}}{x^{2} + 2x - 3} = 2\]
\[ОДЗ:x^{2} + 2x - 8 \neq 0\]
\[D_{1} = 1 + 8 = 9\]
\[x_{1} = - 1 + 3 = 2;\]
\[x_{2} = - 1 - 3 = - 4;\]
\[x \neq 2;\ \ x \neq - 4.\]
\[x^{2} + 2x - 3 \neq 0\]
\[D_{1} = 1 + 3 = 4\]
\[x_{1} = - 1 + 2 = 1;\]
\[x_{2} = - 1 - 2 = - 3.\]
\[x \neq - 3;\ \ x \neq 1.\]
\[\frac{24\left( x^{2} + 2x - 3 \right) - 15\left( x^{2} + 2x - 8 \right)}{(x - 2)(x + 4)(x + 3)(x - 1)} = 2^{(x^{2} + 2x - 3)(x^{2} + 2x - 8)}\]
\[x = 0.\]
\[x = - 2.\]
\[Ответ:x = - 2;x = 0;x = - 1 \pm \sqrt{16,5}.\]
