ГДЗ по алгебре 8 класс Мерзляк Задание 686

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\[1)\ x² + 2x + \frac{3}{x - 8} = \frac{3}{x - 8} + 80\]

\[x^{2} + 2x + \frac{3}{x - 8} - \frac{3}{x - 8} - 80 =\]

\[= 0\]

\[\left\{ \begin{matrix} x^{2} + 2x - 80 = 0 \\ x - 8 \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} + 2x - 80 = 0 \\ x \neq 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[D = 4 + 4 \cdot 80 = 324\]

\[x = \frac{- 2 \pm 18}{2}\]

\[\left\{ \begin{matrix} x = - 10 \\ x = 8\ \ \ \\ x \neq 8\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:\ x = - 10.\]

\[2)\ x² + 8 \cdot \left( \sqrt{x} \right)^{2} - 33 = 0\]

\[\left\{ \begin{matrix} x^{2} + 8x - 33 = 0 \\ x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[D = 64 + 4 \cdot 33 = 196\]

\[x = \frac{- 8 \pm 14}{2}\]

\[\left\{ \begin{matrix} x = 3\ \ \ \ \ \\ x = - 11 \\ x \geq 0\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x = 3.\]