Решебник по алгебре 9 класс Мерзляк Задание 244

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Пояснение.

Решение.

\[1)\ f(x) = \sqrt{x} - 1\]

\[y = \sqrt{x},\ \ E(y) = \lbrack 0; + \infty)\]

\[y = \sqrt{x} - 1,\ \ \]

\[E(y) = \lbrack - 1; + \infty)\]

\[Ответ:\lbrack - 1;\ + \infty).\]

\[2)\ f(x) = 5 - x^{2}\]

\[y = x^{2},\ \ E(y) = \lbrack 0; + \infty)\]

\[y = - x^{2},\ \ E(y) = ( - \infty;0\rbrack\]

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

\[E(y) = ( - \infty;5\rbrack\]

\[Ответ:( - \infty;5\rbrack.\]

\[3)\ f(x) = - 7,\ \ E(f) = - 7\]

\[Ответ:\left\{ - 7 \right\}.\]

\[4)\ f(x) = |x| + 2\]

\[y = |x|,\ \ E(y) = \lbrack 0;\ + \infty)\]

\[y = |x| + 2,\ \ E(y) = \lbrack 2;\ + \infty)\]

\[Ответ:\lbrack 2; + \infty).\]

\[5)\ f(x) = \sqrt{- x^{2}}\]

\[D(f) = - x^{2} \geq 0\]

\[x = 0\]

\[E(f) = 0\]

\[Ответ:\left\{ 0 \right\}.\]

\[6)\ f(x) = \sqrt{x - 2} + \sqrt{2 - x}\]

\[D(f) = \left\{ \begin{matrix} x - 2 \geq 0 \\ 2 - x \geq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x \geq 2 \\ x \leq 2 \\ \end{matrix} \right.\ \]

\[x = 2,\ \ y = 0\]

\[E(f) = 0\]

\[Ответ:\left\{ 0 \right\}\text{.\ \ }\]