Для каждого значения а решите уравнение: (a^2-4a-5)x=a^2-25.

Ответ:

\[\left( a^{2} - 4a - 5 \right)x = a² - 25\]

\[a^{2} - 4a - 5 = 0\]

\[a_{1} + a_{2} = 4;\ \ a_{1} \cdot a_{2} = - 5\]

\[a_{1} = 5,\ \ a_{2} = - 1\]

\[a^{2} - 4a - 5 = (a - 5)(a + 1).\]

\[1)\ \ a = 5:\]

\[\left( 5^{2} - 4 \cdot 5 - 5 \right)x = 5^{2} - 25\ \ \]

\[0 \cdot x = 0\]

\[\Longrightarrow x - любое.\]

\[2)\ a = - 5:\]

\[x = \frac{- 5 + 5}{- 5 + 11} = \frac{0}{- 4} = 0.\]

\[3)\ \ a = - 1:\]

\[нет\ решения.\]

\[4)\ \ a \neq \pm 5;a \neq - 1:\]

\[x = \frac{a + 5}{a + 1}.\]