Решебник по алгебре 8 класс Макарычев ФГОС Задание 548

 

\[\boxed{\text{548\ (549).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x^{2} - 2x - 1 = 0\]

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

\[\left\{ \begin{matrix} y = x^{2}\text{\ \ \ \ \ \ \ \ } \\ y = 2x + 1 \\ \end{matrix} \right.\ \]

\[y = x^{2}\]

\[y = 2x + 1\]

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

\[D = 4 + 4 = 8\]

\[x_{1} = \frac{2 - \sqrt{8}}{2} = \frac{2 - 2\sqrt{2}}{2} = - 0,4;\]

\[x_{2} = \frac{2 + \sqrt{8}}{2} = \frac{2 + 2\sqrt{2}}{2} = 2,4.\]

\[Ответ:x = - 0,4;\ \ x = 2,4.\ \]

\[\textbf{б)}\ x^{2} - 4x + 2 = 0\]

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

\[\left\{ \begin{matrix} y = x^{2} \\ y = 4x - 2 \\ \end{matrix} \right.\ \]

\[y = x^{2}\]

\[y = 4x - 2\]

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

\[D = 16 - 8 = 8\]

\[x_{1} = \frac{4 - \sqrt{8}}{2} = \frac{4 - 2\sqrt{2}}{2} = 0,6;\]

\[x_{2} = \frac{4 + \sqrt{8}}{2} = \frac{4 + 2\sqrt{2}}{2} = 3,4.\]

\(\ \)

\[Ответ:x = 3,4;\ \ x = 0,6.\]

\[\boxed{\text{548.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ x^{2} - 8x + 9 = 0\]

\[D_{1} = 4^{2} - 9 = 16 - 9 = 7\]

\[x_{1} = 4 + \sqrt{7} \approx 6,65;\]

\[x_{2} = 4 - \sqrt{7} \approx 1,35.\]

\[Ответ:x = 6,65;\ \ x = 1,35.\]

\[\textbf{б)}\ 2y^{2} - 8y + 5 = 0\]

\[D_{1} = 4^{2} - 2 \cdot 5 = 16 - 10 = 6\]

\[y_{1} = \frac{4 + \sqrt{6}}{2} \approx 3,22;\ \ \]

\[y_{2} = \frac{4 - \sqrt{6}}{2} \approx 0,78.\ \]

\[Ответ:y = 0,78;\ \ y = 3,22.\]