Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 156

\[\boxed{\mathbf{156}.}\]

\[1)\ \sqrt{5x^{2} + 9x - 2}\]

\[5x^{2} + 9x - 2 \geq 0\]

\[D = 81 + 40 = 121\]

\[x_{1} = \frac{- 9 - 11}{10} = - 2;\ \]

\[\ x_{2} = \frac{- 9 + 11}{10} = 0,2.\]

\[5 \cdot (x + 2)(x - 0,2) \geq 0\]

\[Ответ:x \leq - 2;\ \ x \geq 0,2.\]

\[2)\ \sqrt{- 3x^{2} + x + 4}\]

\[- 3x^{2} + x + 4 \geq 0\]

\[3x^{2} - x - 4 \leq 0\]

\[D = 1 + 48 = 49\]

\[x_{1} = \frac{1 + 7}{6} = \frac{8}{6} = \frac{4}{3} = 1\frac{1}{3};\ \ \]

\[x_{2} = \frac{1 - 7}{6} = - 1.\]

\[3 \cdot (x + 1)\left( x - 1\frac{1}{3} \right) \geq 0\]

\[Ответ:\left\lbrack - 1;1\frac{1}{3} \right\rbrack.\]

\[3)\ \frac{1}{\sqrt{- x^{2} + 2x - 1}}\]

\[- x^{2} + 2x - 1 > 0\]

\[x^{2} - 2x + 1 < 0\]

\[(x - 1)^{2} < 0\]

\[Ответ:нет\ корней.\]

\[4)\ \frac{5}{\sqrt{x^{2} + 6x + 9}}\]

\[x^{2} + 6x + 9 > 0\]

\[(x + 3)^{2} > 0\]

\[Ответ:x - любое\ число,\ \]

\[кроме\ x = - 3.\]