ГДЗ по алгебре 8 класс Мерзляк Задание 755

\[\boxed{\mathbf{755\ (755).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{x^{2} + x - 6}{x + 3} =\]

\[= \frac{(x + 3)(x - 2)}{(x + 3)} = x - 2\]

\[x^{2} + x - 6 = (x + 3)(x - 2)\]

\[x_{1} + x_{2} = - 1,\ \ x_{1} = - 3\]

\[x_{1} \cdot x_{2} = - 6,\ \ x_{2} = 2\]

\[2)\ \frac{x - 4}{x^{2} - 10x + 24} =\]

\[= \frac{(x - 4)}{(x - 6)(x - 4)} = \frac{1}{x - 6}\]

\[x^{2} - 10x + 24 = (x - 6)(x - 4)\]

\[x_{1} + x_{2} = 10,\ \ x_{1} = 6\]

\[x_{1} \cdot x_{2} = 24,\ \ x_{2} = 4\]

\[3)\ \frac{3x - 15}{x^{2} - x - 20} =\]

\[= \frac{3 \cdot (x - 5)}{(x - 5)(x + 4)} = \frac{3}{x + 4}\]

\[x^{2} - x - 20 = (x - 5)(x + 4)\]

\[x_{1} + x_{2} = 1,\ \ x_{1} = 5\]

\[x_{1} \cdot x_{2} = - 20,\ \ x_{2} = - 4\]

\[4)\ \frac{x^{2} - 3x + 2}{6x - 6} =\]

\[= \frac{(x - 2)(x - 1)}{6 \cdot (x - 1)} = \frac{x - 2}{6}\]

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

\[x_{1} + x_{2} = 3,\ \ x_{1} = 2\]

\[x_{1} \cdot x_{2} = 2,\ \ x_{2} = 1\]

\[5)\ \frac{x^{2} - 7x + 12}{x^{2} - 3x} =\]

\[= \frac{(x - 3)(x - 4)}{x(x - 3)} = \frac{x - 4}{x}\]

\[x^{2} - 7x + 12 = (x - 3)(x - 4)\]

\[x_{1} + x_{2} = 7,\ \ x_{1} = 3\]

\[x_{1} \cdot x_{2} = 12,\ \ x_{2} = 4\]

\[6)\ \frac{x^{2} + 4x}{x^{2} + 2x - 8} =\]

\[= \frac{x(x + 4)}{(x + 4)(x - 2)} = \frac{x}{x - 2}\]

\[x^{2} + 2x - 8 = (x + 4)(x - 2)\]

\[x_{1} + x_{2} = - 2,\ \ x_{1} = - 4\]

\[x_{1} \cdot x_{2} = - 8,\ \ x_{2} = 2\]