ГДЗ по алгебре 9 класс Макарычев Задание 301

\[\boxed{\text{301\ (301).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{12 - 5x - 2x^{2}}{15 - 10x}\]

\[- 2x^{2} - 5x + 12 = 0\]

\[2x^{2} + 5x - 12 = 0\]

\[D = 25 + 4 \cdot 2 \cdot 12 = 25 + 9121\]

\[x_{1,2} = \frac{- 5 \pm 11}{4} = - 4;\frac{3}{2};\]

\[12 - 5x - 2x^{2} =\]

\[= - 2 \cdot (x + 4)\left( x - \frac{3}{2} \right) =\]

\[= (x + 4)(3 - 2x);\]

\[\Longrightarrow \frac{12 - 5x - 2x^{2}}{15 - 10x} =\]

\[= \frac{(x + 4)(3 - 2x)}{5 \cdot (3 - 2x)} = \frac{x + 4}{5}.\]

\[\textbf{б)}\ \frac{3x^{2} - 36x - 192}{x^{2} - 256}\]

\[3x^{2} - 36x - 192 = 0\ \ \ \ \ |\ :3\ \]

\[x^{2} - 12x - 64 = 0\]

\[D_{1} = 6^{2} + 64 = 100\]

\[x_{1,2} = 6 \pm 10 = 16;\ - 4;\]

\[3x^{2} - 36x - 192 =\]

\[= 3 \cdot (x - 16)(x + 4);\]

\[\ \Longrightarrow \frac{3x^{2} - 36x - 192}{x^{2} - 256} =\]

\[= \frac{3 \cdot (x - 16)(x + 4)}{(x - 16)(x + 16)} =\]

\[= \frac{3 \cdot (x + 4)}{x + 16} = \frac{3x + 12}{x + 16}.\]