Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1341

\[\boxed{\mathbf{1341}\mathbf{.}}\]

\[\left| x^{2} - 8x + 5 \right| = 2x\]

\[x^{2} - 8x + 5 \geq 0\]

\[D = 64 - 20 = 44\]

\[x_{1} = \frac{8 - \sqrt{44}}{2} \approx \frac{8 - 6,6}{2} \approx\]

\[\approx \frac{1,4}{2} \approx 0,7;\]

\[x_{2} = \frac{8 + \sqrt{44}}{2} \approx \frac{8 + 6,6}{2} \approx\]

\[\approx \frac{14,6}{2} \approx 7,3;\]

\[(x - 0,7)(x - 7,3) \geq 0\]

\[x \leq - 0,7;\ \ x \geq - 7,3.\ \]

\[x \leq - 0,7\ \ и\ \ x \geq - 7,3:\]

\[x^{2} - 8x + 5 = 2x\]

\[x^{2} - 10x + 5 = 0\]

\[D = 100 - 20 = 80\]

\[x = \frac{10 \pm \sqrt{80}}{2} = \frac{10 \pm 4\sqrt{5}}{2} =\]

\[= 5 \pm 2\sqrt{5}.\]

\[- 0,7 < x < - 7,3:\]

\[- \left( x^{2} - 8x + 5 \right) = 2x\]

\[- x^{2} + 8x - 5 - 2x = 0\]

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

\[D = 36 - 20 = 16\]

\[x_{1} = \frac{6 - 4}{2} = 1;\ \]

\[x_{2} = \frac{6 + 4}{2} = 5.\]

\[Ответ:\ \ x = 5.\]