Решебник по алгебре 9 класс Мерзляк Задание 245

 

\[\boxed{\text{245\ (245).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ f(x) = x^{2} + 3\]

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

\[y = x^{2} + 3,\ \ E(y) = \lbrack 3;\ + \infty)\]

\[Ответ:\lbrack 3;\ + \infty).\]

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

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

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

\[y = - \sqrt{x} + 6,\ \ \]

\[E(y) = ( - \infty;6\rbrack\]

\[Ответ:( - \infty;6\rbrack.\]

\[3)\ f(x) = \sqrt{x} \cdot \sqrt{x}\]

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

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

\[Ответ:\lbrack 0;\ + \infty)\text{.\ }\]

\[\boxed{\text{245.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \left( \frac{1^{\backslash a - 8}}{a} + \frac{1^{\backslash a}}{a - 8} \right)\left( (a - {4)}^{\backslash a - 4} - \frac{16}{a - 4} \right) =\]

\[= \frac{a - 8 + a}{a(a - 8)} \cdot \frac{a^{2} - 8a + 16 - 16}{a - 4} =\]

\[= \frac{(2a - 8)\left( a^{2} - 8a \right)}{a(a - 8)(a - 4)} =\]

\[= \frac{2(a - 4)a(a - 8)}{a(a - 8)(a - 4)} = 2.\]

\[Не\ зависит\ от\ переменной.\]

\[2)\ \frac{a}{b - a} - \frac{\text{ac}}{b - c} \cdot \left( \frac{b + c}{bc - ac} - \frac{a + b}{ab - a^{2}} + \frac{b}{\text{ac}} \right) = - 1\]

\[\frac{b + c}{bc - ac} - \frac{a + b}{ab - a^{2}} + \frac{b}{\text{ac}} =\]

\[= \frac{b + c^{\backslash a}}{c(b - a)} - \frac{a + b^{\backslash c}}{a(b - a)} + \frac{b^{\backslash b - a}}{\text{ac}} =\]

\[= \frac{ab + ac - ac - bc + b^{2} - ab}{\text{ac}(b - a)} =\]

\[= \frac{b^{2} - bc}{\text{ac}(b - a)} = \frac{b(b - c)}{\text{ac}(b - a)}\]

\[\frac{\text{ac}}{b - c} \cdot \frac{b(b - c)}{\text{ac}(b - a)} = \frac{b}{b - a}\]

\[\frac{a}{b - a} - \frac{b}{b - a} = \frac{a - b}{b - a} = - 1.\]

\[Не\ зависит\ от\ переменных.\]