Для каждого значения a решите неравенство: x^2-(a-4)x-4a>=0.

Ответ:

\[x^{2} - (a - 4)x - 4a \geq 0\]

\[D = a^{2} - 8a + 16 + 16a =\]

\[= a^{2} + 8a + 16 =\]

\[= (a + 4)^{2} \geq 0 - при\]

\[любом\ значении\ \]

\[переменной\ a.\]

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

\[= - \frac{8}{2} = - 4;\]

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

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

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

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

\[При\ a < - 4:\]

\[x \leq - a;\ \ \ x \geq - 4.\]

\[При\ a = - 4:\]

\[x \in ( - \infty; + \infty).\]

\[При\ a > - 4:\]

\[x \leq - 4;\ \ \ x \geq a.\]