Решебник по алгебре 8 класс Мерзляк ФГОС Задание 1136

\[\boxed{\mathbf{1130}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ (x - 4)(x + 2) - 2 \cdot (3x + 1)(x - 3) =\]

\[= x \cdot (x + 27)\]

\[x^{2} - 2x - 8 - 6x^{2} + 16x + 6 - x^{2} - 27x = 0\]

\[- 6x^{2} - 13x - 2 = 0\]

\[D = 169 - 48 = 121\]

\[x = \frac{13 \pm 11}{- 12}\]

\[x = - 2\]

\[x = - \frac{1}{6}\]

\[Ответ:\ x = - 2;\ x = - \frac{1}{6}.\]

\[2)(4x - 3)^{2} + (3x - 1)(3x + 1) = 9\]

\[16x^{2} - 24x + 9 - 9x^{2} - 1 - 9 = 0\]

\[25x^{2} - 24x - 1 = 0\]

\[D = 576 + 100 = 676\]

\[x = \frac{24 \pm 26}{50}\]

\[x = 1\]

\[x = - \frac{1}{25} = - 0,04\ \]

\[Ответ:\ x = - 0,04;x = 1.\]

\[3)\ \frac{2 \cdot \left( x^{2} - 9 \right)}{5} - \frac{x + 1}{2} = \frac{x - 41}{4}\]

\[16x^{2} - 144 - 20x - 20 - 10x + 410 = 0\]

\[16x² - 30x + 246 = 0\ \ \ \ |\ :2\]

\[8x^{2} - 15x + 123 = 0\]

\[D = 225 - 4 \cdot 8 \cdot 123 < 0\]

\[Ответ:нет\ корней.\]

\[4)\ \frac{x^{2} + 5x}{3} - \frac{x + 3}{2} = \frac{2x^{2} - 2}{8}\]

\[16x² + 80x - 24 - 72 - 12x^{2} + 12 = 0\]

\[4x² + 80x - 84 = 0\ \ \ \ |\ :4\]

\[x² + 20x - 21 = 0\]

\[x_{1} + x_{2} = - 20,\ \ x_{1}x_{2} = - 21,\ \ \]

\[x_{1} = - 21,\ \ x_{2} = 1\]

\[Ответ:\ x = - 21;x = 1.\]