Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1332

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\[1)\ x - 4 + \frac{1}{x} = 0\ \ \ \ \ | \bullet x\]

\[x^{2} - 4x + 1 = 0\]

\[D = 16 - 4 = 12 = 4 \bullet 3\]

\[x = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}.\]

\[Ответ:\ \ x = 2 \pm \sqrt{3}.\]

\[2)\ \frac{4x^{2}}{x + 2} - \frac{10}{x + 2} + 4 = 0\ \ \ \ \ | \bullet (x + 2)\]

\[4x^{2} - 10 + 4(x + 2) = 0\]

\[4x^{2} - 10 + 4x + 8 = 0\]

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

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

\[D = 4 + 8 = 12 = 4 \bullet 3\]

\[x = \frac{- 2 \pm \sqrt{12}}{2 \bullet 2} = \frac{- 2 \pm 2\sqrt{3}}{4} =\]

\[= \frac{- 1 \pm \sqrt{3}}{2}.\]

\[Ответ:\ \ x = \frac{- 1 \pm \sqrt{3}}{2}.\]