Для каждого значения а решите уравнение: x^2+(2-4a)x+3a^2-2a=0.

\[x² + (2 - 4a)x + 3a² - 2a = 0\]

\[= 4a^{2} - 8a + 4 =\]

\[= 4 \cdot \left( a^{2} - 2a + 1 \right) =\]

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

\[при\ \ D = 0:\]

\[(a - 1)^{2} = 0\]

\[a - 1 = 0\]

\[a = 1.\]

\[1)\ \ a = 1:\]

\[x = \frac{4 \cdot 1 - 2 \pm |1 - 1|}{2} = \frac{2}{2} = 1.\]

\[2)\ D > 0;\ \ a > 1:\]

\[x_{1} = \frac{4a - 2 + 2a - 2}{2} =\]

\[= \frac{6a - 4}{2} = 3a - 2;\]

\[x_{2} = \frac{4a - 2 - 2a + 2}{2} = \frac{2a}{2} = a.\]

\[Ответ:\ a = 1 \Longrightarrow \ x = 1;\ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ a > 1\ \Longrightarrow x_{1} = 3a - 2;\ \]

\[x_{2} = a.\]