Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 642

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\[1)\ \sqrt{3 - x} = 2;\]

\[\left( \sqrt{3 - x} \right)^{2} = 2^{2};\]

\[3 - x = 4;\]

\[- x = 1;\]

\[x = - 1;\]

\[Выражение\ имеет\ смысл\ при:\]

\[3 - x \geq 0;\]

\[x \leq 3;\]

\[Ответ:\ \ x = - 1.\]

\[2)\ \sqrt{3x + 1} = 8;\]

\[\left( \sqrt{3x + 1} \right)^{2} = 8^{2};\]

\[3x + 1 = 64;\]

\[3x = 63;\]

\[x = 21;\]

\[Выражение\ имеет\ смысл\ при:\]

\[3x + 1 \geq 0;\]

\[3x \geq - 1;\]

\[x \geq - \frac{1}{3};\]

\[Ответ:\ \ x = 21.\]

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

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

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

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

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

\[= 16 + 48 = 64\]

\[x_{1} = \frac{- 4 - 8}{2 \bullet 4} = - \frac{12}{8} = - 1,5;\]

\[x_{2} = \frac{- 4 + 8}{2 \bullet 4} = \frac{4}{8} = 0,5;\]

\[Выражение\ имеет\ смысл\ при:\]

\[3 - 4x \geq 0;\]

\[4x \leq 3;\]

\[x \leq 0,75;\]

\[Уравнение\ имеет\ решения\ при:\]

\[2x \geq 0;\]

\[x \geq 0;\]

\[Ответ:\ \ x = 0,5.\]

\[4)\ \sqrt{5x - 1 + 3x^{2}} = 3x;\]

\[\left( \sqrt{5x - 1 + 3x^{2}} \right)^{2} = (3x)^{2};\]

\[5x - 1 + 3x^{2} = 9x^{2};\]

\[6x^{2} - 5x + 1 = 0;\]

\[D = 5^{2} - 4 \bullet 6 = 25 - 24 = 1\]

\[x_{1} = \frac{5 - 1}{2 \bullet 6} = \frac{4}{12} = \frac{1}{3};\]

\[x_{2} = \frac{5 + 1}{2 \bullet 6} = \frac{6}{2 \bullet 6} = \frac{1}{2}.\]

\[Ответ:\ \ x = \frac{1}{3};\ \ x = \frac{1}{2}.\]

\[5)\ \sqrt[3]{x^{2} - 17} = 2;\]

\[\left( \sqrt[3]{x^{2} - 17} \right)^{3} = 2^{3};\]

\[x^{2} - 17 = 8;\]

\[x^{2} = 25;\]

\[x = \pm 5;\]

\[Ответ:\ \ x = \pm 5.\]

\[6)\ \sqrt[4]{x^{2} + 17} = 3;\]

\[\left( \sqrt[4]{x^{2} + 17} \right)^{4} = 3^{4};\]

\[x^{2} + 17 = 81;\]

\[x^{2} = 64;\]

\[x = \pm 8;\]

\[Выражение\ имеет\ смысл\ при:\]

\[x^{2} + 17 \geq 0;\]

\[x^{2} \geq - 17 - при\ любом\ x;\]

\[Ответ:\ \ x = \pm 8.\]