ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 650

\[\boxed{\mathbf{650}.}\]

\[1)\ \sqrt{x + 4} - 3\sqrt[4]{x + 4} + 2 = 0;\]

\[Пусть\ y = \sqrt[4]{x + 4}:\]

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

\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]

\[y_{1} = \frac{3 - 1}{2} = 1\ \ и\ \ \]

\[y_{2} = \frac{3 + 1}{2} = 2;\]

\[Первое\ значение:\]

\[\sqrt[4]{x + 4} = 1;\]

\[x + 4 = 1;\]

\[x = - 3;\]

\[Второе\ значение:\]

\[\sqrt[4]{x + 4} = 2;\]

\[x + 4 = 16;\]

\[x = 12;\]

\[Ответ:\ \ x_{1} = - 3;\ \ x_{2} = 12.\]

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

\[Пусть\ y = \sqrt[4]{x - 3}:\]

\[y^{2} = 3y + 4;\]

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

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

\[y_{1} = \frac{3 - 5}{2} = - 1\ \ и\]

\[\text{\ \ }y_{2} = \frac{3 + 5}{2} = 4;\]

\[Первое\ значение:\]

\[\sqrt[4]{x - 3} = - 1 - нет\ корней:\]

\[Второе\ значение:\]

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

\[x - 3 = 256;\]

\[x = 259;\]

\[Ответ:\ \ x = 259.\]

\[3)\ \sqrt[6]{1 - x} - 5\sqrt[3]{1 - x} = - 6;\]

\[Пусть\ y = \sqrt[6]{1 - x}:\]

\[y - 5y^{2} = - 6;\]

\[5y^{2} - y - 6 = 0;\]

\[D = 1^{2} + 4 \bullet 5 \bullet 6 =\]

\[= 1 + 120 = 121\]

\[y_{1} = \frac{1 - 11}{2 \bullet 5} = - 1\ \ и\ \ \]

\[y_{2} = \frac{1 + 11}{2 \bullet 5} = \frac{12}{10} = \frac{6}{5};\]

\[Первое\ значение:\]

\[\sqrt[6]{1 - x} = - 1 - нет\ корней:\]

\[Второе\ значение:\]

\[\sqrt[6]{1 - x} = \frac{6}{5};\]

\[1 - x = \frac{46\ 656}{15\ 625};\]

\[x = \frac{15\ 625}{15\ 625} - \frac{46\ 656}{15\ 625};\]

\[x = - \frac{31\ 031}{15\ 625} = - 1,985984;\]

\[Ответ:\ \ x = - 1,985984.\]

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

\[Пусть\ y = \sqrt{x^{2} + 3x}:\]

\[y^{2} + y = 2;\]

\[y^{2} + y - 2 = 0;\]

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

\[y_{1} = \frac{- 1 - 3}{2} = - 2\ \ и\]

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

\[Первое\ значение:\]

\[\sqrt{x^{2} + 3x} = - 2 - нет\ корней;\]

\[Второе\ значение:\]

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

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

\[x^{2} + 3x - 1 = 0;\]

\[D = 3^{2} + 4 = 9 + 4 = 13\]

\[x = \frac{- 3 \pm \sqrt{13}}{2};\]

\[Ответ:\ \ x = \frac{- 3 \pm \sqrt{13}}{2}.\]

\[5)\ \frac{\sqrt{3 - x} + \sqrt{3 + x}}{\sqrt{3 - x} - \sqrt{3 + x}} = 2;\]

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

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

\[3\sqrt{3 + x} = \sqrt{3 - x};\]

\[9(3 + x) = 3 - x;\]

\[27 + 9x = 3 - x;\]

\[10x = - 24;\]

\[x = - 2,4;\]

\[Выражение\ имеет\ смысл\ при:\]

\[3 - x \geq 0 \Longrightarrow \ \ \ x \leq 3;\]

\[3 + x \geq 0\ \ \Longrightarrow \ \ x \geq - 3;\]

\[Ответ:\ \ x = - 2,4.\]

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

\[+ \sqrt{11 + x - 6\sqrt{x + 2}} = 1;\]

\[\sqrt{x + 2 - 2 \bullet 2\sqrt{x + 2} + 4} +\]

\[+ \sqrt{x + 2 - 2 \bullet 3\sqrt{x + 2} + 9} = 1;\]

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

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

\[\left| \sqrt{x + 2} - 2 \right| + \left| \sqrt{x + 2} - 3 \right| = 1;\]

\[Числа\ под\ знаком\ модуля:\]

\[\sqrt{x + 2} - 2 \geq 0\ \ \Longrightarrow \ \ \]

\[\Longrightarrow \ \sqrt{x + 2} \geq 2\ \Longrightarrow\]

\[\Longrightarrow \ \ \ x + 2 \geq 4 \Longrightarrow \ \ \ x \geq 2;\]

\[\sqrt{x + 2} - 3 \geq 0\ \ \Longrightarrow\]

\[\Longrightarrow \text{\ \ }\sqrt{x + 2} \geq 3\ \Longrightarrow \ \]

\[\Longrightarrow \ \ x + 2 \geq 9\ \ \ \Longrightarrow \ \ \ x \geq 7;\]

\[Если\ x \geq 7:\]

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

\[2\sqrt{x + 2} = 6;\]

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

\[x + 2 = 9;\]

\[x = 7;\]

\[Если\ 2 \leq x < 7:\]

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

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

\[1 = 1;\]

\[x - любое\ допустимое\ число;\]

\[Если\ x < 2:\]

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

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

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

\[4 = 2\sqrt{x + 2};\]

\[2 = \sqrt{x + 2};\]

\[4 = x + 2;\]

\[x = 2;\]

\[Выражение\ имеет\ смысл\ при:\]

\[x + 2 \geq 0;\]

\[x \geq - 2;\]

\[Ответ:\ \ 2 \leq x \leq 7.\]