Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 647

\[\boxed{\mathbf{647}.}\]

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

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

\[x\left( x^{2} - 3x + 2 \right) \geq 0\]

\[x(x - 1)(x - 2) \geq 0\]

\[x \in \lbrack 0;1\rbrack \cup \lbrack 2; + \infty).\]

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

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

\[x^{3} - 2x^{2} - 3x \leq 0\]

\[x\left( x^{2} - 2x - 3 \right) \leq 0\]

\[x(x + 1)(x - 3) \leq 0\]

\[x \in ( - \infty; - 1\rbrack \cup \lbrack 0;3\rbrack.\]

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

\[= \sqrt[5]{\left( x^{3} - x^{3} \right)^{3}}\]

\[x \in R - так\ как\ стпень\ \]

\[корня\ нечетная.\]

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

\[= \frac{1}{\sqrt[4]{\left( x^{4} + x^{3} \right)^{3}}}\]

\[x^{4} + x^{3} > 0\]

\[x^{3}(x + 1) > 0\]

\[x \in ( - \infty; - 1) \cup (0; + \infty).\]

\[5)\ y = \frac{3}{\sqrt[6]{3x^{2} + 14x + 8}}\]

\[3x^{2} + 14x + 8 > 0\]

\[D_{1} = 49 - 24 = 25\]

\[x_{1} = \frac{- 7 - 5}{3} = - 4;\ \ \]

\[x_{2} = \frac{- 7 + 5}{3} = - \frac{2}{3}.\]

\[(x + 4)\left( x + \frac{2}{3} \right) > 0\]

\[x \in ( - \infty; - 4) \cup \left( - \frac{2}{3}; + \infty \right).\]

\[6)\ y = \frac{1}{3\sqrt{x - \sqrt{x + 2}}}\]

\[x - \sqrt{x + 2} > 0\]

\[\sqrt{x + 2} < x\]

\[\left\{ \begin{matrix} x + 2 \geq 0\ \ \ \ \ \ \ \ \ \ \ \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - x - 2 > 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x \geq - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x + 1)(x - 2) > 0 \\ \end{matrix} \right.\ \]

\[x > 2.\]