Решебник по алгебре 9 класс Макарычев Задание 85

 

\[\boxed{\text{85\ (85).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{36 - x^{2}}{6 - 7x + x^{2}}\text{\ \ }\]

\[x^{2} - 7x + 6 = 0\]

\[D = 7^{2} - 4 \cdot 6 = 25\]

\[x_{1} = \frac{7 + 5}{2} = 6;\ \ \ x_{2} = \frac{7 - 5}{2} = 1.\]

\[x^{2} - 7x + 6 = (x - 6)(x - 1);\]

\[\Longrightarrow \frac{36 - x^{2}}{6 - 7x + x^{2}} =\]

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

\[при\ x = - 9:\]

\[\frac{- 9 + 6}{1 + 9} = - \frac{3}{10} = - 0,3;\]

\[при\ x = - 99:\]

\[\frac{- 99 + 6}{1 + 99} = \frac{- 93}{100} = - 0,93;\]

\[при\ x = - 999:\]

\[\frac{- 999 + 6}{1 + 999} = \frac{- 993}{1000} = - 0,993.\]

\[\textbf{б)}\ \frac{4x^{2} + 8x - 32}{4x^{2} - 16} = \frac{x^{2} + 2x - 18}{x^{2} - 4}\]

\[x^{2} + 2x - 8 = 0\]

\[D_{1} = 1 + 8 = 9\]

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

\[= - 1 - 3 = - 4.\]

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

\[\Longrightarrow \frac{4x^{2} + 8x - 32}{4x^{2} - 16} =\]

\[= \frac{x^{2} + 2x - 18}{x^{2} - 4} =\]

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

\[при\ x = - 1:\]

\[\frac{- 1 + 4}{- 1 + 2} = \frac{3}{1} = 3;\]

\[при\ x = 5:\]

\[\frac{5 + 4}{5 + 2} = \frac{9}{7} = 1\frac{2}{7};\]

\[при\ \ x = 10:\]

\[\frac{10 + 4}{10 + 2} = \frac{14}{12} = \frac{7}{6} = 1\frac{1}{6}.\]

\[\boxed{\text{85.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[В\ порядке\ убывания:\]

\[a^{2};\ \frac{1}{a};\ \ a - 2.\]