ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 649

\[\boxed{\mathbf{649}.}\]

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

\[Уравнение\ имеет\ решения\ при:\]

\[\sqrt{x - 3} - \sqrt{2x - 1} \geq 0;\]

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

\[x - 3 \geq 2x - 1;\]

\[- x \geq 2;\]

\[x \leq - 2 - нет\ корней;\]

\[Ответ:\ \ нет\ решений.\]

\[2)\ 2\sqrt{x + 3} - \sqrt{2x + 7} = \sqrt{x};\]

\[2\sqrt{x + 3} = \sqrt{x} + \sqrt{2x + 7};\]

\[4(x + 3) = x +\]

\[+ 2\sqrt{x(2x + 7)} + 2x + 7;\]

\[4x + 12 = 3x + 7 +\]

\[+ 2\sqrt{2x^{2} + 7x};\]

\[x + 5 = 2\sqrt{2x^{2} + 7x};\]

\[x^{2} + 10x + 25 = 4\left( 2x^{2} + 7x \right);\]

\[x^{2} + 10x + 25 = 8x^{2} + 28x;\]

\[7x^{2} + 18x - 25 = 0;\]

\[D = 18^{2} + 4 \bullet 7 \bullet 25 =\]

\[= 324 + 700 = 1024\]

\[x_{1} = \frac{- 18 - 32}{2 \bullet 7} = - \frac{50}{14} =\]

\[= - \frac{25}{7} = - 3\frac{4}{7};\]

\[x_{2} = \frac{- 18 + 32}{2 \bullet 7} = \frac{14}{14} = 1;\]

\[Выполним\ проверку:\]

\[2\sqrt{- \frac{25}{7} + 3} - \sqrt{- 2 \bullet \frac{25}{7} + 7} -\]

\[- \sqrt{- \frac{25}{7}} - не\ имеет\ смысла;\]

\[2\sqrt{1 + 3} - \sqrt{2 + 7} - \sqrt{1} = 2\sqrt{4} -\]

\[- \sqrt{9} - 1 = 4 - 3 - 1 = 0;\]

\[Ответ:\ \ x = 1.\]

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

\[x - 3 = 2x + 1 -\]

\[- 2\sqrt{(2x + 1)(x + 4)} + x + 4;\]

\[2\sqrt{2x^{2} + 8x + x + 4} = 2x + 8;\]

\[\sqrt{2x^{2} + 9x + 4} = x + 4;\]

\[2x^{2} + 9x + 4 = x^{2} + 8x + 16;\]

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

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

\[x_{1} = \frac{- 1 - 7}{2} = - 4\ \ и\ \]

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

\[Выполним\ проверку:\]

\[\sqrt{- 4 - 3}\ldots - не\ имеет\ смысла:\]

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

\[= \sqrt{0} - \sqrt{7} + \sqrt{7} = 0;\]

\[Ответ:\ \ x = 3.\]

\[4)\ \ \sqrt{9 - 2x} = 2\sqrt{4 - x} - \sqrt{1 - x};\]

\[\sqrt{9 - 2x} + \sqrt{1 - x} = 2\sqrt{4 - x};\]

\[9 - 2x + 2\sqrt{(9 - 2x)(1 - x)} +\]

\[+ 1 - x = 4(4 - x);\]

\[10 - 3x +\]

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

\[= 16 - 4x;\]

\[2\sqrt{2x^{2} - 11x + 9} = 6 - x;\]

\[4\left( 2x^{2} - 11x + 9 \right) =\]

\[= 36 - 12x + x^{2};\]

\[8x^{2} - 44x + 36 =\]

\[= 36 - 12x + x^{2};\]

\[7x^{2} - 32x = 0;\]

\[x(7x - 32) = 0;\]

\[x_{1} = 0\ \ и\ \ x_{2} = \frac{32}{7} = 4\frac{4}{7};\]

\[Выполним\ проверку:\]

\[\sqrt{9 - 2 \bullet 4\frac{4}{7}}\ldots - не\ имеет\]

\[\ смысла;\]

\[\sqrt{9 - 2 \bullet 0} - 2\sqrt{4 - 0} + \sqrt{1 - 0} =\]

\[= \sqrt{9} - 2\sqrt{4} + \sqrt{1} = 0;\]

\[Ответ:\ \ x = 0.\]