Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 169

\[\boxed{\mathbf{169.}}\]

\[1)\ \sqrt{2x^{2} + 3x - 2} > 0\]

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

\[D = 3^{2} + 4 \bullet 2 \bullet 2 = 9 + 16 = 25\]

\[x_{1} = \frac{- 3 - 5}{2 \bullet 2} = - \frac{8}{4} = - 2\]

\[x_{2} = \frac{- 3 + 5}{2 \bullet 2} = \frac{2}{4} = 0,5\]

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

\[x < - 2;\text{\ \ }x > 0,5.\]

\[Ответ:\ \ x < - 2;\ \ x > 0,5.\]

\[2)\ \sqrt{2 + x - x^{2}} > - 1;\]

\[Верно\ при\ любом\ допустимом\ \]

\[значении\ x.\]

\[Выражение\ имеет\ смысл\ при:\]

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

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

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

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

\[x_{2} = \frac{1 + 3}{2} = 2.\]

\[(x + 1)(x - 2) \leq 0\]

\[- 1 \leq x \leq 2.\]

\[Ответ:\ \ - 1 \leq x \leq 2.\]

\[3)\ \sqrt{6x - x^{2}} < \sqrt{5}\]

\[6x - x^{2} < 5\]

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

\[D = 6^{2} - 4 \bullet 5 = 36 - 20 = 16\]

\[x_{1} = \frac{6 - 4}{2} = 1;\text{\ \ }x_{2} = \frac{6 + 4}{2} = 5.\]

\[(x - 1)(x - 5) > 0\]

\[x < 1;\text{\ \ }x > 5.\]

\[Выражение\ имеет\ смысл\ при:\]

\[6x - x^{2} \geq 0\]

\[x^{2} - 6x \leq 0\]

\[x(x - 6) \leq 0\]

\[0 \leq x \leq 6.\]

\[Ответ:\ \ 0 \leq x < 1;\ \ 5 < x \leq 6.\]

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

\[x^{2} - x > 2\]

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

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

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

\[x_{2} = \frac{1 + 3}{2} = 2.\]

\[(x + 1)(x - 2) > 0\]

\[x < - 1;\text{\ \ }x > 2.\]

\[Выражение\ имеет\ смысл\ при:\]

\[x^{2} - x \geq 0\]

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

\[x \leq 0;\ \ \ x \geq 1.\]

\[Ответ:\ \ x < - 1;\ \ x > 2.\]

\[5)\ \sqrt{x^{2} + 2x} > - 3 - x^{2}\]

\[\sqrt{x^{2} + 2x} > - \left( x^{2} + 3 \right)\]

\[Верно\ при\ любом\ допустимом\ \]

\[значении\ x.\]

\[Выражение\ имеет\ смысл\ при:\]

\[x^{2} + 2x \geq 0\]

\[(x + 2)x \geq 0\]

\[x \leq - 2;\ \ \ \text{\ \ }x \geq 0.\]

\[Ответ:\ \ x \leq - 2;\ \ x \geq 0.\]

\[6)\ \sqrt{4x - x^{2}} > - 2 - 3x^{2}\]

\[\sqrt{4x - x^{2}} > - \left( 3x^{2} + 2 \right)\]

\[Верно\ при\ любом\ допустимом\ \]

\[значении\ x.\]

\[Выражение\ имеет\ смысл\ при:\]

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

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

\[x(x - 4) \leq 0\]

\[0 \leq x \leq 4.\]

\[Ответ:\ \ 0 \leq x \leq 4.\]