\[\boxed{\text{369\ (369).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ {x^{2}}^{\backslash 4x - 7} = \frac{7x - 4}{4x - 7};\ \ \ \ \ \]
\[4x \neq 7;\ \ x \neq \frac{7}{4}\]
\[4x^{3} - 7x^{2} = 7x - 4\]
\[4x^{3} - 7x^{2} - 7x + 4 = 0\]
\[x = - 1 \rightarrow один\ из\ корней\ \]
\[уравнения.\]
\[(x + 1)\left( 4x^{2} - 11x + 4 \right) = 0\]
\[4x^{2} - 11x + 4 = 0\]
\[D = 121 - 4 \cdot 4 \cdot 4 = 57\]
\[x_{1,2} = \frac{11 \pm \sqrt{57}}{8}.\]
\[Ответ:x = - 1;\ x = \frac{11 \pm \sqrt{57}}{8}.\]
\[\textbf{б)}\ {x^{2}}^{\backslash 3x - 5} = \frac{5x - 3}{3x - 5};\ \ \ \ \ \]
\[3x - 5 \neq 0;\ \ \ x \neq \frac{5}{3}\]
\[3x^{3} - 5x^{2} = 5x - 3\]
\[3x^{3} - 5x^{2} - 5x + 3 = 0\]
\[x = - 1 \rightarrow один\ из\ корней\ \]
\[уравнения.\]
\[(x + 1)\left( 3x^{2} - 8x + 3 \right) = 0\]
\[3x^{2} - 8x + 3 = 0\]
\[D = 16 - 3 \cdot 3 = 7\]
\[x_{1,2} = \frac{4 \pm \sqrt{7}}{3}.\]
\[Ответ:x = - 1;x = \frac{4 \pm \sqrt{7}}{3}.\]
\[\boxed{\text{369.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
Пояснение.
Решение.
\[x^{2} + y^{2} - 6 \cdot (x - y) = 7\]
\[x^{2} + y^{2} - 6x + 6y = 7\]
\[\left( x^{2} - 6x + 9 \right) +\]
\[+ \left( y^{2} + 6y + 9 \right) = 25\]
\[(x - 3)^{2} + (y + 3)^{2} = 25\]
\[Получили\ уравнение\ \]
\[окружности\ с\ центром\ \]
\[в\ точке\ (3; - 3)\ и\]
\[радиусом\ r = 5.\]
\[Значит,\ график\ уравнения\ \]
\[x^{2} + y^{2} - 6 \cdot (x - y) =\]
\[= 7 - окружность.\]
\[Что\ и\ требовалось\ доказать.\]