Для каждого значения а решите уравнение: (a^2-6a-27)x=3a^2+10a+3.

Ответ:

\[\left( a^{2} - 6a - 27 \right)x =\]

\[= 3a² + 10a + 3\]

\[x = \frac{3a^{2} + 10a + 3}{a^{2} - 6a - 27} =\]

\[3a^{2} + 10a + 3 = 0\]

\[D_{1} = 25 - 9 = 16\]

\[a_{1} = \frac{- 5 + 4}{3} = - \frac{1}{3};\ \ \ \ \]

\[a_{2} = \frac{- 5 - 4}{3} = - 3.\]

\[3a^{2} + 10a + 3 =\]

\[= 3 \cdot \left( a + \frac{1}{3} \right)(a + 3) =\]

\[= (3a + 1)(a + 3).\]

\[a^{2} - 6a - 27 = 0\]

\[a_{1} + a_{2} = 6;\ \ a_{1} \cdot a_{2} = - 27\]

\[a_{1} = 9;\ \ \ \ a_{2} = - 3\]

\[a^{2} - 6a - 27 = (a - 9)(a + 3).\]

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

\[\left( ( - 3)^{2} - 6 \cdot ( - 3) - 27 \right) \cdot x =\]

\[= 3 \cdot ( - 3)^{2} + 10 \cdot ( - 3) + 3.\]

\[0 \cdot x = 0\]

\[x - любое\ число.\]

\[2)\ \ a = 9:\]

\[нет\ решения.\]

\[3)\ \ a = - \frac{1}{3}:\]

\[x = 0.\]

\[4)\ \ a \neq - 3,\ a \neq 9,\ a \neq - \frac{1}{3}:\]

\[x = \frac{3a + 1}{a - 9}.\]