ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 466

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\[1)\ \frac{\sqrt{(x + 3)^{2} - 12x}}{\sqrt[4]{x^{3}} - \frac{3}{\sqrt[4]{x}}} =\]

\[= \frac{\sqrt{x^{2} + 6x + 9 - 12x}}{\frac{x - 3}{\sqrt[4]{x}}} =\]

\[= \frac{\sqrt[4]{x} \cdot \sqrt{(x - 3)^{2}}}{(x - 3)} =\]

\[= \frac{\sqrt[4]{x} \cdot |x - 3|}{(x - 3)} = \pm \sqrt[4]{x}\]

\[x + y = \sqrt{2a + 2a - 4} =\]

\[= 2\sqrt{a - 1}.\]

\[A = \frac{2\sqrt{a - 1}}{a - 2};\]

\[1 < a < 2:\]

\[A = \frac{2}{2 - a}.\]

\[t = \sqrt{x + \frac{2xy}{1 + y^{2}}};\ \ \]

\[k = \sqrt{x - \frac{2xy}{1 + y^{2}}};\]

\[B = t + k;\]

\[t^{2} + k^{2} = x + \frac{2xy}{1 + y^{2}} +\]

\[+ x - \frac{2xy}{1 + y^{2}} = 2x;\]

\[B^{2} = (t + k)^{2} = t^{2} + 2tk + k^{2} =\]

\[B^{2} = 2x + 2 \cdot \frac{x\left| 1 - y^{2} \right|}{1 + y^{2}}\]

\[1)\ 1 - y^{2} \leq 0\]

\[y^{2} \geq 0:\]

\[B^{2} = \frac{4xy^{2}}{1 + y^{2}};\]

\[\textbf{а)}\ y \leq - 1:\]

\[B = - 2\sqrt{x};\]

\[\textbf{б)}\ y \geq 1:\]

\[A = 2\sqrt{x}.\]

\[2)\ 1 - y^{2} > 0\]

\[y^{2} < 1:\]

\[B^{2} = \frac{4x}{1 + y^{2}};\ \ \ \ B = \frac{2\sqrt{x}}{\sqrt{1 + y^{2}}}\]

\[\textbf{а)} - 1 < y < 0:\]

\[A = - \frac{2\sqrt{x}}{y}.\]

\[\textbf{б)}\ 0 < y < 1:\]

\[A = \frac{2\sqrt{x}}{y}\text{.\ }\]