Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по алгебре 8 класс Мерзляк ФГОС Задание 791
Задание 791
\[\boxed{\mathbf{791\ (791).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ \frac{2x - 10}{x³ + 1} + \frac{4}{x + 1} = \frac{5x - 1}{x² - x + 1}\]
\[- x^{2} - 6x - 5 = 0\]
\[x^{2} + 6x + 5 = 0\]
\[x_{1} + x_{2} = - 6,\ \ x_{1}x_{2} = 5,\ \ \]
\[x_{1} = - 5,\ \ \]
\[x_{2} = - 1\ (не\ подходит)\]
\[Ответ:\ x = - 5.\]
\[2)\ \frac{6}{x^{2} - 4x + 3} + \frac{5 - 2x}{x - 1} = \frac{3}{x - 3}\]
\[x^{2} - 4x + 3 = (x - 3)(x - 1)\]
\[x_{1} + x_{2} = 4,\ \ x_{1}x_{2} = 3,\ \ \]
\[\text{\ \ }x_{1} = 3,\ \ x_{2} = 1\]
\[- 2x^{2} + 8x - 6 = 0\ \ \ \ \ |\ :( - 2)\]
\[x^{2} - 4x + 3 \neq 0\]
\[Ответ:корней\ нет.\]
\[3)\ \frac{4x - 6}{x + 2} - \frac{x}{x + 1} = \frac{14}{x^{2} + 3x + 2}\]
\[x^{2} + 3x + 2 = (x + 2)(x + 1)\]
\[x_{1} + x_{2} = - 3,\ \ x_{1}x_{2} = 2,\]
\[\text{\ \ }x_{1} = - 2,\ \ x_{2} = - 1\]
\[3x^{2} - 4x - 20 = 0\]
\[D = 16 + 240 = 256\]
\[x = \frac{4 - 16}{6} = - 2\ (не\ подходит)\]
\[x = \frac{4 + 16}{6} = \frac{10}{3} = 3\frac{1}{3}\]
\[Ответ:x = 3\frac{1}{3}.\]
\[4)\ \frac{x}{x^{2} - 4} - \frac{3x - 1}{x^{2} + x - 6} =\]
\[= \frac{2}{x² + 5x + 6}\]
\[x^{2} + x - 6 = (x + 6)(x - 2)\]
\[x_{1} + x_{2} = - 1,\ \ x_{1}x_{2} = - 6,\ \ \]
\[x_{1} = - 3,\ \ x_{2} = 2\]
\[x^{2} + 5x + 6 = (x + 3)(x + 2)\]
\[x_{1} + x_{2} = - 5,\ \ x_{1}x_{2} = 6,\ \ \]
\[x_{1} = - 3,\ \ x_{2} = - 2\]
\[x \neq 2;\ \ \ x \neq - 2;\ \ x \neq - 3\]
\[- 2x^{2} - 4x + 6 = 0\ \ \ \ |\ :( - 2)\]
\[x² + 2x - 3 = 0\]
\[x_{1} + x_{2} = - 2,\ \ x_{1}x_{2} = - 3,\]
\[\text{\ \ }x_{1} = - 3\ (не\ подходит),\ \ \]
\[x_{2} = 1\]
\[Ответ:x = 1.\]
\[\boxed{\mathbf{7}\mathbf{9}\mathbf{1}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ x² - 3x - 18 = 0\]
\[x_{1} + x_{2} = 3,\ \ x_{1} = - 3\]
\[x_{1} \cdot x_{2} = - 18,\ \ x_{2} = 6\]
\[x² - 3x - 18 = (x + 3)(x - 6)\]
\[2)\ x² + 5x - 14 = 0\]
\[x_{1} + x_{2} = - 5,\ \ x_{1} = - 7\]
\[x_{1} \cdot x_{2} = 14,\ \ x_{2} = 2\]
\[x^{2} + 5x - 14 = (x + 7)(x - 2)\]
\[3) - x^{2} + 3x + 4 = 0\]
\[x_{1} + x_{2} = 3,\ \ x_{1} = 4\]
\[x_{1} \cdot x_{2} = - 4,\ \ x_{2} = - 1\]
\[- x^{2} + 3x + 4 =\]
\[= - (x - 4)(x + 1) =\]
\[= (4 - x)(x + 1)\]
\[4)\ 5x² + 8x - 4 = 0\]
\[x_{1} + x_{2} = - \frac{8}{5},\ \ \]
\[x_{1} = - \frac{10}{5} = - 2\ \]
\[x_{1} \cdot x_{2} = - \frac{4}{5},\ \ x_{2} = \frac{2}{5}\ \]
\[5x^{2} + 8x - 4 =\]
\[= 5 \cdot (x + 2)\left( x - \frac{2}{5} \right) =\]
\[= (x + 2)(5x - 2)\]
\[5)\ 2a² - 3a + 1 = 0\]
\[a_{1} + a_{2} = \frac{3}{2},\ \ a_{1} = \frac{2}{2} = 1\]
\[a_{1} \cdot a_{2} = \frac{1}{2},\ \ a_{2} = \frac{1}{2}\]
\[2a^{2} - 3a + 1 =\]
\[= 2 \cdot (a - 1)\left( a - \frac{1}{2} \right) =\]
\[= (a - 1)(2a - 1)\]
\[6)\ 4b² - 11b - 3 = 0\]
\[b_{1} + b_{2} = \frac{11}{4},\ \ b_{1} = - \frac{1}{4}\]
\[b_{1} \cdot b_{2} = - \frac{3}{4},\ \ b_{2} = \frac{12}{4} = 3\]
\[4b^{2} - 11b - 3 =\]
\[= 4 \cdot \left( b + \frac{1}{4} \right)(b - 3) =\]
\[= (4b + 1)(b - 3)\]
\[7) - \frac{1}{4}x^{2} - 2x - 3 = 0\]
\[x_{1} + x_{2} = - 8,\ \ x_{1} = - 2\]
\[x_{1} \cdot x_{2} = 12,\ \ x_{2} = - 6\]
\[- \frac{1}{4}x^{2} - 2x - 3 =\]
\[= - \frac{1}{4} \cdot (x + 2)(x + 6)\]
\[8)\ 0,3m² - 3m + 7,5 = 0\]
\[m_{1} + m_{2} = \frac{3}{0,3} = 10,\ \ m_{1} = 5\]
\[m_{1} \cdot m_{2} = \frac{7,5}{0,3} = 25,\ \ m_{2} = 5\]
\[0,3m^{2} - 3m + 7,5 =\]
\[= 0,3 \cdot (m - 5)²\]
\[9)\ x² - 2x - 2 = 0\]
\[D = 4 + 8 = 12\]
\[x = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}\]
\[x^{2} - 2x - 2 =\]
\[= (x - 1 - \sqrt{3})(x + 1 - \sqrt{3})\]
