Решебник по алгебре 9 класс Макарычев Задание 84

 

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

Пояснение.

Решение.

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

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

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

\[= 121 - 96 = 25\]

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

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

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

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

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

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

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

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

\[= 81 + 40 = 121\]

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

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

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

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

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

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

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

\[\boxed{\text{84.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\frac{123}{23} = 5\frac{8}{23} \rightarrow D\left( \frac{123}{23} \right).\]

\[\left( 1\frac{2}{3} \right)^{2} = \left( \frac{5}{3} \right)^{2} = \frac{25}{9} =\]

\[= 2\frac{7}{9} \rightarrow B\left( 2\frac{7}{9} \right).\]

\[(0,8)^{- 1} = \left( \frac{4}{5} \right)^{- 1} = \frac{5}{4} =\]

\[= 1\frac{1}{4} \rightarrow A\left( 1\frac{1}{4} \right).\]