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МатематикаГДЗ по алгебре 8 класс Мерзляк Задание 778
Задание 778
\[\boxed{\mathbf{778\ (778).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ \frac{x^{2} - 5x - 6}{x - 6} = 0;\ \ \ \ x \neq 6\]
\[x^{2} - 5x - 6 = 0\]
\[x_{1} + x_{2} = 5,\ \ x_{1}x_{2} = - 6,\ \]
\[\ x_{1} = 6\ (не\ подходит),\ \ \]
\[x_{2} = - 1\ \]
\[Ответ:\ x = - 1.\]
\[2)\ \frac{4x^{2} - 7x - 2}{x - 2} = 0;\ \ \ \ \ \ \ x \neq 2\]
\[4x^{2} - 7x - 2 = 0\]
\[D = 49 + 32 = 81,\ \ \]
\[x_{1,2} = \frac{7 \pm 9}{8}\]
\[x_{1} = 2\ (не\ подходит);\ \ \ \ \ \]
\[x_{2} = - \frac{1}{4} = - 0,25\]
\[Ответ:\ x = - 0,25.\]
\[3)\ \frac{2x² + 6}{x + 8} = \frac{13x}{x + 8}\ \]
\[\frac{2x^{2} + 6 - 13x}{x + 8} = 0;\ \ \ \ \ \ \ x \neq - 8\]
\[2x^{2} - 13x + 6 = 0\]
\[\ \ D = 169 - 48 = 121,\ \]
\[\ x_{1,2} = \frac{13 \pm 11}{4}\]
\[x_{1} = 6;\ \ \ \ \ x_{2} = 0,5\]
\[Ответ:x = 6;x = 0,5.\]
\[4)\ \frac{x² + 4x}{x + 7} = \frac{5x + 56}{x + 7}\]
\[\frac{x^{2} + 4x - 5x - 56}{x + 7} = 0;\ \ \ \ \ \]
\[x \neq - 7\]
\[x^{2} - x - 56 = 0\]
\[x_{1} + x_{2} = 1,\ \ x_{1}x_{2} = - 56,\ \ \]
\[x_{1} = - 7\ (не\ подходит),\ \ \]
\[x_{2} = 8\ \]
\[Ответ:x = 8.\]
\[5)\ \frac{x² + 12x}{x + 4} - \frac{5x - 12}{x + 4} = 0\]
\[\frac{x^{2} + 12x - 5x + 12}{x + 4} = 0;\ \ \ \]
\[x \neq 0\]
\[x^{2} + 7x + 12 = 0\]
\[x_{1} + x_{2} = - 7,\ \ x_{1}x_{2} = 12,\ \ \]
\[x_{1} = - 3,\ \ \]
\[x_{2} = - 4\ (не\ подходит)\]
\[Ответ:\ x = - 3.\]
\[6)\ \frac{x^{2} - 3x}{x + 6} = 6\]
\[\frac{x^{2} - 3x}{x + 6} - 6 = 0;\ \ \ \ \ \ \ x \neq - 6\]
\[x^{2} - 3x - 6 \cdot (x + 6) = 0\]
\[x^{2} - 3x - 6x - 36 = 0\]
\[x^{2} - 9x - 36 = 0\]
\[\ x_{1} + x_{2} = 9,\ \ x_{1}x_{2} = - 36,\ \ \]
\[x_{1} = - 3,\ \ x_{2} = 12\ \]
\[Ответ:\ x = - 3;x = 12.\]
\[7)\ \frac{2 - 33y}{y - 4} = 7y\ \ \ \ \ \ \ \ \ | \cdot (y - 4);\ \ \ \ \ \ \ \]
\[y \neq 4\]
\[2 - 33y - 7y \cdot (y - 4) = 0\]
\[2 - 33y - 7y^{2} + 28y = 0\]
\[- 7y^{2} - 5y + 2 = 0\]
\[{7y}^{2} + 5y - 2 = 0\]
\[\ \ D = 25 + 56 = 81\]
\[\text{\ \ }y_{1,2} = \frac{5 \pm 9}{- 14}\]
\[y_{1} = - 1;\ \ \ y_{2} = \frac{2}{7}\]
\[Ответ:\ y = - 1;y = \frac{2}{7}.\]
\[8)\ y - \frac{39}{y} = 10\ \ \ \ \ \ \ \ \ | \cdot y \neq 0\]
\[y^{2} - 39 - 10y = 0\]
\[y^{2} - 10y - 39 = 0\]
\[y_{1} + y_{2} = 10,\ \ y_{1}y_{2} = - 39,\ \ \]
\[y_{1} = - 3,\ \ y_{2} = 13\]
\[Ответ:\ y = - 3;y = 13.\]
