Решите уравнение: x^2-3x-1+3/(x^2-3x+3)=0.

\[x^{2} - 3x - 1 + \frac{3}{x^{2} - 3x + 3} = 0\]

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

\[t - 4 + \frac{3}{t} = 0\ \ \ \ \ \ \ \ \ | \cdot t\]

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

\[D = ( - 4)^{2} - 4 \cdot 1 \cdot 3 =\]

\[= 16 - 12 = 4;\ \ \ \sqrt{D} = 2.\]

\[t_{1} = \frac{4 + 2}{2} = \frac{6}{2} = 3;\ \ \ \]

\[\text{\ \ \ }t_{2} = \frac{4 - 2}{2} = \frac{2}{2} = 1\]

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

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

\[x(x - 3) = 0\]

\[x = 0;\ \ \ \ \ \ \ \ x = 3.\]

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

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

\[D = ( - 3)^{2} - 4 \cdot 1 \cdot 2 = 9 - 8 =\]

\[= 1;\ \ \ \ \sqrt{D} = 1.\]

\[x_{1} = \frac{3 + 1}{2} = \frac{4}{2} = 2;\ \ \ \ \]

\[\text{\ \ \ }x_{2} = \frac{3 - 1}{2} = \frac{2}{2} = 1\]

\[Ответ:0;3;2;1.\]