Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 18

\[\boxed{\mathbf{18}\mathbf{.}}\]

\[\textbf{а)}\ \frac{|x - 2| - |x|}{|2x - 1| + |x|} =\]

\[= \frac{|2x - 1| - |x|}{|x - 2| + |x|}\]

\[x \in R.\]

\[(x - 2)^{2} - x^{2} = (2x - 1)^{2} - x^{2}\]

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

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

\[x^{2} = 3\]

\[x = \pm 1.\]

\[Ответ:x = \pm 1.\]

\[\textbf{б)}\ \frac{|2x - 3| - |x|}{|3x - 2| + |x|} =\]

\[= \frac{|3x - 2| - |x|}{|2x - 3| + |x|}\]

\[x \in R.\]

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

\[4x^{2} - 12x + 9 = 9x^{2} - 12x + 4\]

\[5x^{2} = 5\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[Ответ:x = \pm 1.\]

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

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

\[1)\ x \geq 0.\]

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

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

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

\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = - 4\]

\[x_{1} = 4;\]

\[x_{2} = - 1 < 0\ (не\ подходит).\]

\[Ответ:x = 4.\]

\[\textbf{г)}\ \frac{\sqrt{2x + 4} - \sqrt{x}}{\sqrt{x^{2} + 3} + \sqrt{x}} =\]

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

\[1)\ x \geq 0.\]

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

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

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

\[D_{1} = 1 + 1 = \sqrt{2}\]

\[x_{1} = 1 + \sqrt{2};\]

\[x_{2} = 1 - \sqrt{2} < 0\ (не\ подходит).\]

\[Ответ:x = 1 + \sqrt{2}.\]