Решебник по алгебре 9 класс Макарычев Задание 236

 

\[\boxed{\text{236\ (236).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ f(x) = |x^{2} - 2x|\]

\[Нули\ функции:\]

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

\[x(x - 2) = 0\]

\[x = 0;\ \ \ \ x = 2.\]

\[При\ x < 0:\]

\[f(x) = x^{2} - 2x;\]

\[При\ 0 \leq x \leq 2:\]

\[f(x) = - x^{2} + 2x;\]

\[При\ x > 2:\]

\[f(x) = x^{2} - 2x.\]

\[\]

\[\textbf{б)}\ f(x) = x^{2} - 2|x|\]

\[При\ x < 0:\]

\[f(x) = x^{2} + 2x;\]

\[При\ x \geq 0:\]

\[f(x) = x^{2} - 2x.\]

\[\boxed{\text{236.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{3x - 2}{x - 1} - \frac{2x + 3}{x + 3} = \frac{12x + 4}{x^{2} + 2x - 3}\]

\[\frac{x^{2} - 6x - 7}{(x - 1)(x + 3)} = 0\]

\[\frac{(x - 7)(x + 1)}{(x - 1)(x + 3)} = 0\]

\[(x - 7)(x + 1) = 0\]

\[\Longrightarrow x_{1} = 7,\ \ x_{2} = - 1.\]

\[Ответ:x = 7;\ \ x = - 1.\]

\[\frac{3x^{2} - 32x + 52}{(x - 3)(x + 7)} = 0\]

\[3x^{2} - 32x + 52 = 0\]

\[ОДЗ:\ \ x \neq - 7;3.\]

\[D = 16^{2} - 3 \cdot 52 = 100\]

\[x_{1,2} = \frac{16 \pm 10}{3} = 2;8\frac{2}{3}.\]

\[Ответ:x = 2;\ \ x = 8\frac{2}{3}.\]

\[\textbf{в)}\frac{x}{x^{2} + 4x + 4} =\]

\[= \frac{4}{x^{2} - 4} - \frac{16}{x^{3} + 2x^{2} - 4x - 8}\]

\[\frac{x}{(x + 2)^{2}} =\]

\[= \frac{4}{x^{2} - 4} - \frac{16}{(x + 2)\left( x^{2} - 4 \right)}\]

\[\frac{x(x - 2) - 4 \cdot (x + 2) + 16}{(x + 2)^{2}(x - 2)} = 0\]

\[\frac{x^{2} - 6x + 8}{(x + 2)^{2}(x - 2)} = 0\]

\[\frac{(x - 2)(x - 4)}{(x + 2)^{2}(x - 2)} = 0\]

\[\frac{x - 4}{(x + 2)^{2}} = 0\]

\[x - 4 = 0\]

\[x = 4.\]

\[Ответ:x = 4.\]