Решебник по алгебре 9 класс Макарычев Задание 913

\[\boxed{\text{913\ (913).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[= \frac{(7 - m - 2) \cdot 2}{m - 3} = \frac{10 - 2m}{m - 3}\]

\[= \frac{a + 5}{a - 3} \cdot \frac{a^{2} - 3a + 9}{a^{2} + 3a - 10} =\]

\[= \frac{a + 5}{a - 3} \cdot \frac{a^{2} - 3a + 9}{(a - 2)(a + 5)} =\]

\[= \frac{a² - 3a + 9}{(a - 2)(a - 3)} = \frac{a² - 3a + 9}{a² - 5a + 6}\]

\[a^{2} + 3a - 10 = (a - 2)(a + 5)\]

\[x_{1} + x_{2} = - 3;\ \ \ \ x_{1} \cdot x_{2} = - 10\]

\[x_{1} = 2;\ \ \ \ x_{2} = - 5.\]

\[= \frac{2x^{2} - 14x + 20}{(x - 2)(x + 2)(x - 5)} =\]

\[= \frac{2 \cdot (x - 5)(x - 2)}{(x - 2)(x + 2)(x - 5)} = \frac{2}{x + 2}\]

\[2x^{2} - 14x + 20 = (x - 5)(x - 2)\]

\[D = 196 - 160 = 36\]

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

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

\[= \frac{6 - 2x}{3 - x} = \frac{2 \cdot (3 - x)}{3 - x} = 2.\]