Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1340

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\[\left| x^{2} - 3x - 6 \right| = 2x\]

\[x^{2} - 3x - 6 \geq 0\]

\[D = 9 + 24 = 33\]

\[x_{1} = \frac{3 - \sqrt{33}}{2} \approx \frac{3 - 5,7}{2} \approx\]

\[\approx - \frac{2,7}{2} \approx - 1,3;\]

\[x_{2} = \frac{3 + \sqrt{33}}{2} \approx \frac{3 + 5,7}{2} \approx\]

\[\approx \frac{8,7}{2} \approx 4,3;\]

\[(x + 1,3)(x - 4,3) \geq 0\]

\[x \leq - 1,3;\ \text{\ \ }x \geq 4,3.\ \]

\[x \leq - 1,3\ \ и\ \ x \geq 4,3:\]

\[x^{2} - 3x - 6 = 2x\]

\[x^{2} - 5x - 6 = 0\]

\[D = 25 + 24 = 49\]

\[x_{1} = \frac{5 - 7}{2} = - 1;\]

\[x_{2} = \frac{5 + 7}{2} = 6.\]

\[- 1,3 < x < 4,3:\]

\[- \left( x^{2} - 3x - 6 \right) = 2x\]

\[- x^{2} + 3x + 6 - 2x = 0\]

\[x^{2} - x - 6 = 0\]

\[D = 1 + 24 = 25\]

\[x_{1} = \frac{1 - 5}{2} = - 2;\]

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

\[Ответ:\ \ x = 3.\]