Решебник по алгебре 9 класс Макарычев Задание 292

 

\[\boxed{\text{292}\text{\ (292)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{a + 1^{\backslash a + 1}}{a - 2} + \frac{a - 4^{\backslash a - 2}}{a + 1} =\]

\[= \frac{3a + 3}{a^{2} - a - 2}\]

\[a^{2} - a - 2 = (a + 1)(a - 2)\]

\[a_{1} + a_{2} = 1;\ \ \ \ a_{1} \cdot a_{2} = - 2\]

\[a_{1} = 2;\ \ \ a_{2} = - 1.\]

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

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

\[D = 7^{2} - 4 \cdot 2 \cdot 6 = 49 - 48 = 1\]

\[a_{1} = \frac{7 - 1}{4} = 1,5;\ \ a_{2} = \frac{7 - 1}{4} = 2.\]

\[Ответ:при\ a = 1,5.\]

\[\textbf{б)}\ \frac{3a - 5}{a^{2} - 1} - \frac{6a - 5}{a - a^{2}} = \frac{3a + 2}{a^{2} + a}\]

\[\frac{3a - 5^{\backslash a}}{(a - 1)(a + 1)} - \frac{6a - 5^{\backslash a + 1}}{a(1 - a)} =\]

\[= \frac{3a + 2^{\backslash a - 1}}{a(a + 1)}\]

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

\[6a^{2} - 3a - 3 = 0\ \ \ \ \ \ |\ :3\]

\[2a^{2} - a - 1 = 0\]

\[D = 1 + 4 \cdot 2 = 9\]

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

\[\text{\ \ }a_{2} = \frac{1 - 3}{4} = - \frac{1}{2}.\]

\[Ответ:a = - 0,5.\]

\[\boxed{\text{292.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

Пояснение.

Решение.

\[\mathbf{Выражение\ под\ знаком\ корня\ }\]

\[\mathbf{должно\ быть\ }\]

\[\mathbf{неотрицательным\ }\left( \mathbf{\geq 0} \right)\mathbf{.}\]

\[\textbf{а)}\ y = \sqrt{(5 - x)(x + 8)}\]

\[(5 - x)(x + 8) \geq 0\]

\[(x + 8)(x - 5) \leq 0\]

\[x \in \lbrack - 8;5\rbrack.\]

\[\textbf{б)}\ y = \sqrt{(x + 12)(x - 1)(x - 9)}\]

\[(x + 12)(x - 1)(x - 9) \geq 0\]

\[x \in \lbrack - 12;1\rbrack \cup \lbrack 9;\ + \infty).\ \]