ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 699

\[\boxed{\mathbf{699}.}\]

\[1)\ (x - 3)^{x^{2} - x - 2} = 1\]

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

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

\[x_{1} = 2;\ \ \ x_{2} = - 1.\]

\[x - 3 = 1\]

\[x = 4.\]

\[Ответ:x = - 1;2;4.\]

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

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

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

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

\[x_{1} = 2;\ \ \ x_{2} = - 1.\]

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

\[x^{2} = 1\]

\[x = \pm 1.\]

\[Ответ:x = \pm 1;2.\]

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

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

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

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

\[x_{1} = - 4;\ \ \ x_{2} = 1.\]

\[2)\ x + 3 = \pm 1\]

\[x = - 4;\ \ x = - 3;\ \ \ x = - 2.\]

\[Ответ:x = - 4;\ - 3;\ - 2;1.\]

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

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

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

\[D_{1} = 1 + 3 = 4\]

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

\[\text{\ \ }x_{2} = 1 - 2 = - 1.\]

\[2)\ x + 3 = 1\]

\[x = - 2.\]

\[Ответ:x = - 2;\ - 1;\ 3.\]