ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 159

\[\boxed{\mathbf{159}.}\]

\[1)\ y = \frac{2x - 1}{x^{2} - 5x + 6}\]

\[x^{2} - 5x + 6 = 0\]

\[x_{1} + x_{2} = 5;\ \ x_{1} \cdot x_{2} = 6\]

\[x_{1} = 2;\ \ x_{2} = 3.\]

\[ОДЗ:\ \ x \neq 2;\ \ x \neq 3.\]

\[2)\ y = \frac{5 - x}{2x^{2} + 3x - 2}\]

\[2x^{2} + 3x - 2 = 0\]

\[D = 9 + 16 = 25\]

\[x_{1} = \frac{- 3 + 5}{4} = \frac{1}{2};\ \ \]

\[x_{2} = \frac{- 3 - 5}{4} = - 2.\]

\[ОДЗ:\ \ x \neq - 2;\ \ x \neq \frac{1}{2}.\]

\[3)\ y = \sqrt{3x + 1}\]

\[3x + 1 \geq 0\]

\[3x \geq - 1\]

\[x \geq - \frac{1}{3}.\]

\[ОДЗ:\ \ x \geq - \frac{1}{3}.\]

\[4)\ y = \sqrt{7 - 3x}\]

\[7 - 3x \geq 0\]

\[- 3x \geq - 7\]

\[x \leq \frac{7}{3}\]

\[ОДЗ:\ \ x \leq 2\frac{1}{3}.\]

\[5)\ y = \frac{3}{\sqrt{x - 5}}\]

\[x - 5 > 0\]

\[x > 5.\]

\[ОДЗ:\ \ x > 5.\]

\[6)\ y = \frac{12}{\sqrt{8 + x}}\]

\[8 + x > 0\]

\[x > - 8\]

\[ОДЗ:\ \ x > - 8.\]

\[7)\ y = \sqrt{x} + \sqrt{3 - x}\]

\[\left\{ \begin{matrix} x \geq 0\ \ \ \ \ \ \ \\ 3 - x \geq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x \geq 0 \\ x \leq 3 \\ \end{matrix} \right.\ \]

\[ОДЗ:\ \ \ 0 \leq x \leq 3.\]

\[8)\ y = \sqrt{x - 7} - \sqrt{x}\]

\[\left\{ \begin{matrix} x - 7 \geq 0 \\ x \geq 0\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x \geq 7 \\ x \geq 0 \\ \end{matrix} \right.\ \]

\[ОДЗ:\ \ x \geq 7.\]