Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 155

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\[1)\ \sqrt{x} - x = - 12\]

\[\sqrt{x} = x - 12\]

\[x = (x - 12)^{2}\]

\[x = x^{2} - 24x + 144\]

\[x^{2} - 25x + 144 = 0\]

\[D = 25^{2} - 4 \bullet 144 =\]

\[= 625 - 576 = 49\]

\[x_{1} = \frac{25 - 7}{2} = 9;\text{\ \ }\]

\[x_{2} = \frac{25 + 7}{2} = 16;\]

\[Уравнение\ имеет\ решения\ при:\]

\[x - 12 \geq 0\]

\[x \geq 12.\]

\[Ответ:\ \ x = 16.\]

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

\[\sqrt{x} = 2x - 2 - x\]

\[\sqrt{x} = x - 2\]

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

\[x = x^{2} - 4x + 4\]

\[x^{2} - 5x + 4 = 0\]

\[D = 5^{2} - 4 \bullet 4 = 25 - 16 = 9\]

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

\[Уравнение\ имеет\ решения\ при:\]

\[x - 2 \geq 0\]

\[x \geq 2\]

\[Ответ:\ \ x = 4.\]

\[3)\ \sqrt{x - 1} = x - 3\]

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

\[x - 1 = x^{2} - 6x + 9\]

\[x^{2} - 7x + 10 = 0\]

\[D = 7^{2} - 4 \bullet 10 = 49 - 40 = 9\]

\[x_{1} = \frac{7 - 3}{2} = 2;\text{\ \ }x_{2} = \frac{7 + 3}{2} = 5.\]

\[Уравнение\ имеет\ решения\ при:\]

\[x - 3 \geq 0\]

\[x \geq 3.\]

\[Ответ:\ \ x = 5.\]

\[4)\ \sqrt{6 + x - x^{2}} = 1 - x\]

\[6 + x - x^{2} = (1 - x)^{2}\]

\[6 + x - x^{2} = 1 - 2x + x^{2}\]

\[2x^{2} - 3x - 5 = 0\]

\[D = 3^{2} + 4 \bullet 2 \bullet 5 = 9 + 40 = 49\]

\[x_{1} = \frac{3 - 7}{2 \bullet 2} = - \frac{4}{4} = - 1;\]

\[x_{2} = \frac{3 + 7}{2 \bullet 2} = \frac{10}{4} = 2,5.\]

\[Выражение\ имеет\ смысл\ при:\]

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

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

\[D = 1^{2} + 4 \bullet 6 = 1 + 24 = 25\]

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

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

\[(x + 2)(x - 3) \leq 0\]

\[- 2 \leq x \leq 3.\]

\[Уравнение\ имеет\ решения\ при:\]

\[1 - x \geq 0\]

\[x \leq 1.\]

\[Ответ:\ \ x = - 1.\]