Для каждого значения a решите неравенство: x^2-(a+3)x+3a<=0.

Ответ:

\[x^{2} - (a + 3)x + 3a \leq 0\]

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

\[D = a^{2} + 6a + 9 - 12a =\]

\[= a^{2} - 6a + 9 =\]

\[= (a - 3)^{2} \geq 0 - для\ всех\ a.\]

\[x_{1} = \frac{(a + 3) - (a - 3)}{2} = \frac{6}{2} = 3;\]

\[x_{2} = \frac{(a + 3) + (a - 3)}{2} =\]

\[= \frac{2a}{2} = a.\]

\[\left( x - x_{1} \right)\left( x - x_{2} \right) \leq 0\]

\[x_{1} \leq x \leq x_{2}.\]

\[Если\ a < 3:\ \]

\[x \in \lbrack a;3\rbrack.\]

\[Если\ \ \ a > 3:\]

\[x \in \lbrack 3;a\rbrack.\]

\[Если\ a = 3:\]

\[x = 3.\]