Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 154

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\[1)\ x + 1 = \sqrt{1 - x};\]

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

\[x^{2} + 2x + 1 = 1 - x;\]

\[x^{2} + 3x = 0;\]

\[x(x + 3) = 0;\]

\[x_{1} = 0;\ \ \ \ x_{2} = - 3.\]

\[Выражение\ имеет\ смысл\ при:\]

\[1 - x \geq 0\]

\[x \leq 1.\]

\[Уравнение\ имеет\ решения\ при:\]

\[x + 1 \geq 0\]

\[x \geq - 1.\]

\[Ответ:\ \ x = 0.\]

\[2)\ x = 1 + \sqrt{x + 11}\]

\[x - 1 = \sqrt{x + 11}\]

\[(x - 1)^{2} = x + 11\]

\[x^{2} - 2x + 1 = x + 11\]

\[x^{2} - 3x - 10 = 0\]

\[D = 3^{2} + 4 \bullet 10 = 9 + 40 = 49\]

\[x_{1} = \frac{3 - 7}{2} = - 2;\text{\ \ }\]

\[x_{2} = \frac{3 + 7}{2} = 5.\]

\[Выражение\ имеет\ смысл\ при:\]

\[x + 11 \geq 0\]

\[x \geq - 11.\]

\[Уравнение\ имеет\ решения\ при:\]

\[x - 1 \geq 0\]

\[x \geq 1.\]

\[Ответ:\ \ x = 5.\]

\[3)\ \sqrt{x + 3} = \sqrt{5 - x}\]

\[x + 3 = 5 - x\]

\[2x = 2\]

\[x = 1.\]

\[Ответ:\ \ x = 1.\]

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

\[x^{2} - x - 3 = 9\]

\[x^{2} - x - 12 = 0\]

\[D = 1^{2} + 4 \bullet 12 = 1 + 48 = 49\]

\[x_{1} = \frac{1 - 7}{2} = - 3;\ \text{\ \ }\]

\[x_{2} = \frac{1 + 7}{2} = 4;\]

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