Для каждого значения а решите уравнение: x^2+(1-5a)x+4a^2-a=0.

\[x² + (1 - 5a)x + 4a² - a = 0\]

\[= 9a^{2} - 6a + 1 = (3a - 1)^{2}\]

\[1)\ При\ D = 0:\]

\[3a - 1 = 0\]

\[3a = 1\ \ \]

\[a = \frac{1}{3}.\]

\[Получаем:\]

\[x = \frac{- (1 - 5a)}{2} = \frac{5a - 1}{2} =\]

\[= \frac{5 \cdot \frac{1}{3} - 1}{2} = \frac{\frac{5}{3} - 1\ }{2} = \frac{\frac{2}{3}}{2} = \frac{1}{3}.\]

\[2.\ \ При\ D > 0;\ a \neq \frac{1}{3}:\]

\[x_{1,2} = \frac{- (1 - 5a) \pm \sqrt{(3a - 1)^{2}}}{2} =\]

\[= \frac{5a - 1 \pm (3a - 1)}{2}\]

\[a > \frac{1}{3}:\]

\[x_{1,2} = \frac{5a - 1 \pm (3a - 1)}{2}\]

\[x_{1} = \frac{5a - 1 + 3a - 1}{2} =\]

\[= \frac{8a - 2}{2} = 4a - 1\]

\[x_{2} = \frac{5a - 1 - 3a + 1}{2} = \frac{2a}{2} = a.\]

\[a < \frac{1}{3}:\]

\[x_{1,2} = \frac{5a - 1 \pm \left( - (3a - 1 \right))\ }{2} =\]

\[= \frac{5a - 1 \pm (1 - 3a)}{2}\]

\[x_{1} = \frac{5a - 1 + 1 - 3a}{2} = \frac{2a}{2} = 2\]

\[x_{2} = \frac{5a - 1 - 1 + 3a}{2} =\]

\[= \frac{8a - 2}{2} = 4a - 1.\]

\[Ответ:если\ a = \frac{1}{3};то\ x = \frac{1}{3};\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ если\ a \neq \frac{1}{3};\]

\[то\ x_{1} = 4a - 1\ или\ \ x_{2} = a.\]