Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 179

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

\[- \infty < x < + \infty\]

\[D(x) = ( - \infty;\ + \infty).\]

\[2)\ y = \sqrt[6]{2 - x^{2}}\]

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

\[x^{2} \leq 2\]

\[- \sqrt{2} \leq x \leq \sqrt{2}\]

\[D(x) = \left\lbrack - \sqrt{2};\ \sqrt{2} \right\rbrack.\]

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

\[3x^{2} + 1 \neq 0\]

\[3x^{2} \neq - 1\]

\[x^{2} \neq - \frac{1}{3} \Longrightarrow при\ любом\ x\]

\[D(x) = ( - \infty;\ + \infty).\]

\[4)\ y = \sqrt{x^{2} - x - 2}\]

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

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

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

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

\[x \leq - 1;\text{\ \ }x \geq 2\]

\[D(x) = ( - \infty;\ - 1\rbrack \cup \lbrack 2;\ + \infty).\]