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МатематикаРешебник по алгебре 8 класс Мерзляк ФГОС Задание 792
Задание 792
\[\boxed{\mathbf{792\ (792).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ \frac{3x + 2}{x^{2} + 2x + 4} + \frac{x^{2} + 39}{x^{3} - 8} =\]
\[= \frac{5}{x - 2}\]
\[- x^{2} - 14x + 15 = 0\]
\[x^{2} + 14x - 15 = 0\]
\[x_{1} + x_{2} = - 14,\ \ x_{1}x_{2} = - 15,\ \]
\[\ x_{1} = - 15,\ \ x_{2} = 1\]
\[Ответ:\ x = - 15;x = 1.\]
\[2)\ \frac{x}{x - 1} + \frac{x + 1}{x + 3} = \frac{8}{x^{2} + 2x - 3}\]
\[x^{2} + 2x - 3 = 0\]
\[x_{1} + x_{2} = - 2,\ \ x_{1}x_{2} = - 3,\]
\[\ x_{1} = - 3,\ \ x_{2} = 1\]
\[x^{2} + 3x + x^{2} - 1 - 8 = 0\]
\[2x^{2} + 3x - 9 = 0\]
\[D = 9 + 72 = 81\]
\[x = \frac{- 3 - 9}{4} = - 3\]
\[x = \frac{- 3 + 9}{4} = 1,5\]
\[Ответ:x = 1,5.\]
\[\boxed{\mathbf{7}\mathbf{9}\mathbf{2}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ \frac{x^{2} + x - 6}{x + 3} =\]
\[= \frac{(x + 3)(x - 2)}{(x + 3)} = x - 2\]
\[x^{2} + x - 6 = (x + 3)(x - 2)\]
\[x_{1} + x_{2} = - 1,\ \ x_{1} = - 3\]
\[x_{1} \cdot x_{2} = - 6,\ \ x_{2} = 2\]
\[2)\ \frac{x - 4}{x^{2} - 10x + 24} =\]
\[= \frac{(x - 4)}{(x - 6)(x - 4)} = \frac{1}{x - 6}\]
\[x^{2} - 10x + 24 = (x - 6)(x - 4)\]
\[x_{1} + x_{2} = 10,\ \ x_{1} = 6\]
\[x_{1} \cdot x_{2} = 24,\ \ x_{2} = 4\]
\[3)\ \frac{3x - 15}{x^{2} - x - 20} =\]
\[= \frac{3 \cdot (x - 5)}{(x - 5)(x + 4)} = \frac{3}{x + 4}\]
\[x^{2} - x - 20 = (x - 5)(x + 4)\]
\[x_{1} + x_{2} = 1,\ \ x_{1} = 5\]
\[x_{1} \cdot x_{2} = - 20,\ \ x_{2} = - 4\]
\[4)\ \frac{x^{2} - 3x + 2}{6x - 6} =\]
\[= \frac{(x - 2)(x - 1)}{6 \cdot (x - 1)} = \frac{x - 2}{6}\]
\[x^{2} - 3x + 2 = (x - 2)(x - 1)\]
\[x_{1} + x_{2} = 3,\ \ x_{1} = 2\]
\[x_{1} \cdot x_{2} = 2,\ \ x_{2} = 1\]
\[5)\ \frac{x^{2} - 7x + 12}{x^{2} - 3x} =\]
\[= \frac{(x - 3)(x - 4)}{x(x - 3)} = \frac{x - 4}{x}\]
\[x^{2} - 7x + 12 = (x - 3)(x - 4)\]
\[x_{1} + x_{2} = 7,\ \ x_{1} = 3\]
\[x_{1} \cdot x_{2} = 12,\ \ x_{2} = 4\]
\[6)\ \frac{x^{2} + 4x}{x^{2} + 2x - 8} =\]
\[= \frac{x(x + 4)}{(x + 4)(x - 2)} = \frac{x}{x - 2}\]
\[x^{2} + 2x - 8 = (x + 4)(x - 2)\]
\[x_{1} + x_{2} = - 2,\ \ x_{1} = - 4\]
\[x_{1} \cdot x_{2} = - 8,\ \ x_{2} = 2\]
