Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 183

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

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

\[3 - x = 4\]

\[- x = 1\]

\[x = - 1\]

\[Выражение\ имеет\ смысл\ при:\]

\[3 - x \geq 0\]

\[x \leq 3.\]

\[Ответ:\ \ x = - 1.\]

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

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

\[3x + 1 = 64\]

\[3x = 63\]

\[x = 21.\]

\[Выражение\ имеет\ смысл\ при:\]

\[3x + 1 \geq 0\]

\[3x \geq - 1\]

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

\[Ответ:\ \ x = 21.\]

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

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

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

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

\[D = 4^{2} + 4 \bullet 4 \bullet 3 = 16 + 48 = 64\]

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

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

\[Выражение\ имеет\ смысл\ при:\]

\[3 - 4x \geq 0\]

\[4x \leq 3\]

\[x \leq 0,75.\]

\[Уравнение\ имеет\ решения\ при:\]

\[2x \geq 0\]

\[x \geq 0.\]

\[Ответ:\ \ x = 0,5.\]

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

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

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

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

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

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

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

\[Выполним\ проверку:\]

\[\sqrt{5 \bullet \frac{1}{3} - 1 + 3 \bullet \frac{1}{9}} - 3 \bullet \frac{1}{3} =\]

\[= \sqrt{\frac{5}{3} - \frac{3}{3} + \frac{1}{3}} - 1 = \sqrt{1} - 1 = 0;\]

\[\sqrt{5 \bullet \frac{1}{2} - 1 + 3 \bullet \frac{1}{4}} - 3 \bullet \frac{1}{2} =\]

\[= \sqrt{\frac{10}{4} - \frac{4}{4} + \frac{3}{4}} - \frac{3}{2} =\]

\[= \sqrt{\frac{9}{4}} - \frac{3}{2} = 0.\]

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

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

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

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

\[x^{2} = 25\]

\[x = \pm 5.\ \]

\[Ответ:\ \ x = \pm 5.\]

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

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

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

\[x^{2} = 64\]

\[x = \pm 8.\]

\[Выражение\ имеет\ смысл\ при:\]

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

\[x^{2} \geq - 17 - при\ любом\ \text{x.}\]

\[Ответ:\ \ x = \pm 8.\]

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