ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 187

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

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

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

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

\[- x \geq 2\]

\[x \leq - 2 - нет\ корней.\]

\[Ответ:\ \ нет\ решений.\]

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

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

\[4(x + 3) =\]

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

\[4x + 12 = 3x + 7 + 2\sqrt{2x^{2} + 7x}\]

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

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

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

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

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

\[= 324 + 700 = 1024\]

\[x_{1} = \frac{- 18 - 32}{2 \bullet 7} = - \frac{50}{14} = - \frac{25}{7} =\]

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

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

\[Проверим:\]

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

\[= 2\sqrt{4} - \sqrt{9} - 1 = 4 - 3 - 1 = 0.\]

\[Ответ:\ \ x = 1.\]

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

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

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

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

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

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

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

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

\[Проверим:\]

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

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

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

\[Ответ:\ \ x = 3.\]

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

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

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

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

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

\[8x^{2} - 44x + 36 = 36 - 12x + x^{2}\]

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

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

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

\[Проверим:\]

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

\[смысла;\]

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

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

\[Ответ:\ \ x = 0.\]