Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 157

\[\boxed{\mathbf{157}.}\]

\[1)\ \frac{x^{2} - 6x + 8}{x^{2} + 2x - 3} \leq 0\]

\[x^{2} - 6x + 8 = (x - 2)(x - 4)\]

\[D_{1} = 9 - 8 = 1\]

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

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

\[D_{1} = 1 + 3 = 4\]

\[x_{1} = - 1 + 2 = 1;\ \]

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

\[\frac{(x - 2)(x - 4)}{(x + 3)(x - 1)} \leq 0\]

\[Ответ:\ - 3 < x < 1;\ \ 2 \leq x \leq 4.\]

\[2)\ \frac{x^{2} + x - 6}{x^{2} + x + 1} \geq 0\]

\[x^{2} + x - 6 = (x + 3)(x - 2)\]

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

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

\[x^{2} + x + 1 = 0\]

\[D = 1 - 4 =\]

\[= - 3 < 0\ (нет\ корней).\]

\[\frac{(x + 3)(x - 2)}{x^{2} + x + 1} \geq 0\]

\[Ответ:x \leq - 3;x \geq 2.\]

\[3)\ \frac{(x + 2)^{2}}{(x - 1)(x + 5)} \geq 0\]

\[Ответ:x < - 5;\ \ x = - 2;\ \ x > 1.\]

\[4)\ \frac{(x + 3)^{2}(x + 1)}{x - 4} \leq 0\]

\[Ответ:x = - 3;\ - 1 \leq x < 4.\]

\[5)\ \frac{(x - 4)^{2}(x + 2)}{(x - 5)^{3}} \leq 0\]

\[Ответ:\ - 2 \leq x < 5.\]

\[6)\ \frac{(x + 7)^{3}(x - 2)}{x + 3} \geq 0\ \]

\[Ответ:\ - 7 \leq x < - 3;\ \ x \geq 2.\]