ГДЗ по алгебре 9 класс Макарычев Задание 369

\[\boxed{\text{369\ (369).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ {x^{2}}^{\backslash 4x - 7} = \frac{7x - 4}{4x - 7};\ \ \ \ \ \]

\[4x \neq 7;\ \ x \neq \frac{7}{4}\]

\[4x^{3} - 7x^{2} = 7x - 4\]

\[4x^{3} - 7x^{2} - 7x + 4 = 0\]

\[x = - 1 \rightarrow один\ из\ корней\ \]

\[уравнения.\]

\[(x + 1)\left( 4x^{2} - 11x + 4 \right) = 0\]

\[4x^{2} - 11x + 4 = 0\]

\[D = 121 - 4 \cdot 4 \cdot 4 = 57\]

\[x_{1,2} = \frac{11 \pm \sqrt{57}}{8}.\]

\[Ответ:x = - 1;\ x = \frac{11 \pm \sqrt{57}}{8}.\]

\[\textbf{б)}\ {x^{2}}^{\backslash 3x - 5} = \frac{5x - 3}{3x - 5};\ \ \ \ \ \]

\[3x - 5 \neq 0;\ \ \ x \neq \frac{5}{3}\]

\[3x^{3} - 5x^{2} = 5x - 3\]

\[3x^{3} - 5x^{2} - 5x + 3 = 0\]

\[x = - 1 \rightarrow один\ из\ корней\ \]

\[уравнения.\]

\[(x + 1)\left( 3x^{2} - 8x + 3 \right) = 0\]

\[3x^{2} - 8x + 3 = 0\]

\[D = 16 - 3 \cdot 3 = 7\]

\[x_{1,2} = \frac{4 \pm \sqrt{7}}{3}.\]

\[Ответ:x = - 1;x = \frac{4 \pm \sqrt{7}}{3}.\]