Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1615

\[\boxed{\mathbf{1615}\mathbf{.}}\]

\[\frac{7 - 3x + \sqrt{x^{2} + 3x - 4}}{x - 3} < - 1\]

\[x - 3 > 0\]

\[x > 3.\]

\[x > 3:\]

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

\[x^{2} + 3x - 4 < 4x^{2} - 16x + 16\]

\[3x^{2} - 19x + 20 > 0\]

\[D = 361 - 240 = 121\]

\[x_{1} = \frac{19 - 11}{2 \bullet 3} = \frac{8}{6} = \frac{4}{3};\]

\[x_{2} = \frac{19 + 11}{2 \bullet 3} = 5;\]

\[\left( x - \frac{4}{3} \right)(x - 5) > 0\]

\[x < \frac{4}{3}\text{\ \ }и\ \ x > 5.\]

\[x < 3:\]

\[7 - 3x + \sqrt{x^{2} + 3x - 4} > - 1 \bullet (x - 3)\]

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

\[2x - 4 \geq 0\]

\[x - 2 \geq 0\]

\[x \geq 2.\]

\[2 \leq x < 3:\]

\[x^{2} + 3x - 4 > 4x^{2} + 16x + 16\]

\[3x^{2} - 19x + 20 < 0\]

\[\left( x - \frac{4}{3} \right)(x - 5) < 0\]

\[\frac{4}{3} < x < 5.\]

\[x < 2:\]

\[при\ любом\ значении\ x.\]

\[Имеет\ смысл\ при:\]

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

\[D = 9 + 16 = 25\]

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

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

\[(x + 4)(x - 1) \geq 0\]

\[x \leq - 4\ \ и\ \ x \geq 1.\]

\[Ответ:\ \ \]

\[x \in ( - \infty;\ - 4\rbrack \cup \lbrack 1;\ 3) \cup (5;\ + \infty).\]