Решебник по алгебре 9 класс Мерзляк Задание 280

 

\[\boxed{\text{280\ (280).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[D(f) = \lbrack a;2\rbrack;\ \text{\ \ }a < 2\]

\(\ \)

\[Если\ a < - 2:\ \]

\[f_{наиб} = f(a) = a^{2};\ \]

\[\ f_{наим} = f(0) = 0.\]

\[Если\ a = - 2:\ \]

\[f_{наиб} = f( - 2) = 4;\ \]

\[\ f_{наим} = f(0) = 0.\]

\[Если - 2 \leq a < 0:\ \]

\[f_{наиб} = f( - 2) = 4;\]

\[\text{\ \ }f_{наим} = f(0) = 0.\]

\[Если\ 0 < a \leq 2:\ \]

\[f_{наиб} = f(2) = 4;\ \ \]

\[f_{наим} = f(a) = a^{2}.\ \]

\[\boxed{\text{280.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ 4x + a = 2\]

\[4x = 2 - a\]

\[x = \frac{2 - a}{4}\]

\[\frac{2 - a}{4} > 0\ \ \ | \cdot 4\]

\[2 - a > 0\]

\[a < 2\]

\[Ответ:при\ a \in ( - \infty;2).\]

\[2)\ (a + 6) \cdot x = 3\]

\[ax + 6x = 3\]

\[x = \frac{3}{a + 6},\ \ a \neq - 6\]

\[\frac{3}{a + 6} < 0\]

\[a + 6 < 0\]

\[a < - 6\]

\[Ответ:при\ a \in ( - \infty;\ - 6).\]

\[3)\ (a - 1) \cdot x = a^{2} - 1\ \]

\[x = \frac{a^{2} - 1}{a - 1}\]

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

\[x = a + 1,\ \ a \neq 1\]

\[a + 1 > 0\]

\[a > - 1\]

\[Ответ:при\ a \in ( - 1;\ + \infty);\ \ \]

\[кроме\ 1.\]