\[\boxed{\text{245\ (245).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[1)\ f(x) = x^{2} + 3\]
\[y = x^{2},\ \ E(y) = \lbrack 0;\ + \infty)\]
\[y = x^{2} + 3,\ \ E(y) = \lbrack 3;\ + \infty)\]
\[Ответ:\lbrack 3;\ + \infty).\]
\[2)\ f(x) = 6 - \sqrt{x}\]
\[y = \sqrt{x},\ \ E(y) = \lbrack 0;\ + \infty)\]
\[y = - \sqrt{x},\ \ E(y) = ( - \infty;0\rbrack\]
\[y = - \sqrt{x} + 6,\ \ \]
\[E(y) = ( - \infty;6\rbrack\]
\[Ответ:( - \infty;6\rbrack.\]
\[3)\ f(x) = \sqrt{x} \cdot \sqrt{x}\]
\[f(x) = \sqrt{x^{2}}\]
\[f(x) = |x|,\ \ E(f) = \lbrack 0; + \infty)\]
\[Ответ:\lbrack 0;\ + \infty)\text{.\ }\]