Найдите решения уравнения (x+3x/(x-3))^2=4-(3x^2)/(x-3).

Ответ:

\[\left( x^{\backslash x - 3} + \frac{3x}{x - 3} \right)^{2} = 4 - \frac{3x^{2}}{x - 3}\]

\[\left( \frac{x^{2} - 3x + 3x}{x - 3} \right)^{2} = 4 - 3\frac{x^{2}}{x - 3}\]

\[\left( \frac{x^{2}}{x - 3} \right)^{2} = 4 - 3\frac{x^{2}}{x - 3}\]

\[y = \frac{x^{2}}{x - 3}:\]

\[y^{2} = 4 - 3y\]

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

\[D = 9 + 16 = 25\]

\[y_{1} = \frac{- 3 + 5}{2} = 1;\ \ y_{2} = \frac{- 3 - 5}{2} = - 4\]

\[1)\ y = - 4:\]

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

\[x^{2} + 4x - 12 = 0\]

\[D = 4 + 12 = 16\]

\[x_{1} = - 2 + 4 = 2;\ \ x_{2} = - 2 - 4 = - 6.\]

\[2)\ y = 1:\]

\[\frac{x^{2}}{x - 3} = 1\]

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

\[D = 1 - 12 < 0\]

\[нет\ корней.\]

\[Ответ:x = 2;x = - 6.\]