Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 716

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

\[1)\ \left\{ \begin{matrix} x^{2} - x + 1 > 1\ \ \\ x^{2} - \frac{5}{2}x + 1 < 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x^{2} - x > 0\ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{2} - 5x + 2 < 0 \\ \end{matrix} \right.\ \]

\[2x^{2} - 5x + 2 = 0\]

\[D = 25 - 16 = 9\]

\[x_{1} = \frac{5 + 3}{4} = 2;\ \ x_{2} = \frac{5 - 3}{4} = \frac{1}{2}.\]

\[\left\{ \begin{matrix} x(x - 1) > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2 \cdot (x - 2)(x - 0,5) < 0 \\ \end{matrix} \right.\ \]

\[1 < x < 2.\]

\[2)\ \left\{ \begin{matrix} 0 < x^{2} - x + 1 < 1 \\ x^{2} - \frac{5}{2}x + 1 > 0\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x^{2} - x + 1 > 0\ \ \ \ \ \\ x^{2} - x < 0\ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{2} - 5x + 2 > 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x(x - 1) < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2 \cdot (x - 2)(x - 0,5) > 0 \\ \end{matrix} \right.\ \]

\[0 < x < 0,5.\]