Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 190

\[\boxed{\mathbf{190}\mathbf{.}}\]

\[1)\ \frac{x^{2} - 13x + 40}{\sqrt{19x - x^{2} - 78}} \leq 0\]

\[x^{2} - 13x + 40 \leq 0\]

\[D = 13^{2} - 4 \bullet 40 =\]

\[= 169 - 160 = 9\]

\[x_{1} = \frac{13 - 3}{2} = 5;\text{\ \ }\]

\[x_{2} = \frac{13 + 3}{2} = 8;\]

\[(x - 5)(x - 8) \leq 0\]

\[5 \leq x \leq 8.\]

\[Выражение\ имеет\ смысл\ при:\]

\[19x - x^{2} - 78 > 0\]

\[x^{2} - 19x + 78 < 0\]

\[D = 19^{2} - 4 \bullet 78 = 361 - 312 =\]

\[= 49\]

\[x_{1} = \frac{19 - 7}{2} = 6;\ \ \ \ \ \]

\[x_{2} = \frac{19 + 7}{2} = 13;\]

\[(x - 6)(x - 13) < 0\]

\[6 < x < 13.\]

\[Ответ:\ \ 6 < x \leq 8.\]

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

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

\[4\left( 2x^{2} + 7x - 4 \right) < (x + 4)^{2}\]

\[8x^{2} + 28x - 16 < x^{2} + 8x + 16\]

\[7x^{2} + 20x - 32 < 0\]

\[D = 20^{2} + 4 \bullet 7 \bullet 32 =\]

\[= 400 + 896 = 1296\]

\[x_{1} = \frac{- 20 - 36}{2 \bullet 7} = - \frac{56}{14} = - 4;\]

\[x_{2} = \frac{- 20 + 36}{2 \bullet 7} = \frac{16}{14} = \frac{8}{7} = 1\frac{1}{7}.\]

\[(x + 4)\left( x - 1\frac{1}{7} \right) < 0\]

\[4 < x < 1\frac{1}{7}\]

\[Выражение\ имеет\ смысл\ при:\]

\[2x^{2} + 7x - 4 \geq 0\]

\[D = 7^{2} + 4 \bullet 2 \bullet 4 =\]

\[= 49 + 32 = 81\]

\[x_{1} = \frac{- 7 - 9}{2 \bullet 2} = - \frac{16}{4} = - 4;\]

\[x_{2} = \frac{- 7 + 9}{2 \bullet 2} = \frac{2}{4} = 0,5.\]

\[(x + 4)(x - 0,5) \geq 0\]

\[x \leq - 4;\ \ \ \text{\ \ }x \geq 0,5.\]

\[Неравенство\ всегда\ верно\ при:\]

\[x + 4 < 0\]

\[x < - 4.\]

\[Ответ:\ \ x < - 4;\ \ 0,5 \leq x < 1\frac{1}{7}.\]

\[3)\ \sqrt{3 + x} > |x - 3|\]

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

\[3 + x > x^{2} - 6x + 9\]

\[x^{2} - 7x + 6 < 0\]

\[D = 7^{2} - 4 \bullet 6 = 49 - 24 = 25\]

\[x_{1} = \frac{7 - 5}{2} = 1;\text{\ \ }x_{2} = \frac{7 + 5}{2} = 6;\]

\[(x - 1)(x - 6) < 0\]

\[1 < x < 6.\]

\[Выражение\ имеет\ смысл\ при:\]

\[3 + x \geq 0.\]

\[x \geq - 3.\]

\[Ответ:\ \ 1 < x < 6.\]

\[4)\ \sqrt{3 - x} < \sqrt{7 + x} + \sqrt{10 + x}\]

\[5x^{2} + 16x - 84 < 0\]

\[D = 16^{2} + 4 \bullet 5 \bullet 84 =\]

\[= 256 + 1680 = 1936\]

\[x_{1} = \frac{- 16 - 44}{2 \bullet 5} = - 6;\text{\ \ }\]

\[x_{2} = \frac{- 16 + 44}{2 \bullet 5} = 2,8.\]

\[(x + 6)(x - 2,8) < 0\]

\[- 6 < x < 2,8.\]

\[Выражение\ имеет\ смысл\ при:\]

\[3 - x \geq 0\ \ \Longrightarrow x \leq 3;\]

\[7 + x \geq 0\ \ \ \Longrightarrow \ \ x \geq - 7;\]

\[10 + x \geq 0\ \ \Longrightarrow \ \ \ x \geq - 10.\]

\[Неравенство\ всегда\ верно\ при:\]

\[- 3x - 14 \leq 0\]

\[3x + 14 \geq 0\]

\[3x \geq - 14\]

\[x \geq - 4\frac{2}{3}.\]

\[Ответ:\ \ - 6 < x \leq 3.\]

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