ГДЗ по алгебре 9 класс Макарычев Задание 84

\[\boxed{\text{84\ (84).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{x^{2} - 11x + 24}{x^{2} - 64}\]

\[x^{2} - 11x + 24 = 0\]

\[D = 11^{2} - 4 \cdot 24 =\]

\[= 121 - 96 = 25\]

\[x_{1} = \frac{11 + 5}{2} = 8;\ \ \ \ x_{2} =\]

\[= \frac{11 - 5}{2} = 3.\]

\[x^{2} - 11x + 24 = (x - 8)(x - 3).\]

\[\Longrightarrow \frac{x^{2} - 11x + 24}{x^{2} - 64} =\]

\[= \frac{(x - 8)(x - 3)}{(x - 8)(x + 8)} = \frac{x - 3}{x + 8}.\]

\[\textbf{б)}\ \frac{2y^{2} + 9y - 5}{4y^{2} - 1}\]

\[2y^{2} + 9y - 5 = 0\]

\[D = 9^{2} + 4 \cdot 2 \cdot 5 =\]

\[= 81 + 40 = 121\]

\[y_{1} = \frac{- 9 - 11}{4} = - 5;\ \ y_{2} =\]

\[= \frac{- 9 + 11}{4} = \frac{1}{2}.\]

\[2y^{2} + 9y - 5 =\]

\[= 2 \cdot (y + 5)\left( y - \frac{1}{2} \right) =\]

\[= (y + 5)(2y - 1).\]

\[\Longrightarrow \frac{2y^{2} + 9y - 5}{4y^{2} - 1} =\]

\[= \frac{(y + 5)(2y - 1)}{(2y - 1)(2y + 1)} = \frac{y + 5}{2y + 1}.\]