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

Ответ:

\[x² - (a - 2)x - 2a \geq 0\]

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

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

\[= (a + 2)^{2} \geq 0 \Longrightarrow верно\ при\ \]

\[любом\ a.\]

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

\[= - \frac{4}{25} = - 2;\]

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

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

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

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

\[При\ a < - 2:\]

\[x \in ( - \infty;a\rbrack \cup \lbrack - 2; + \infty).\]

\[При\ a = - 2:\]

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

\[При\ a > - 2:\]

\[x \in ( - \infty; - 2\rbrack \cup \lbrack a; + \infty).\]