Для каждого значения а решите уравнение: (a^2-9a-10)x=3a^2+a-2.

\[\left( a^{2} - 9a - 10 \right)x = 3a² + a - 2\]

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

\[a_{1} + a_{2} = - \frac{1}{3};\ \ a_{1} \cdot a_{2} = - \frac{2}{3}\]

\[a_{1} = \frac{2}{3},\ \ a_{2} = - 1.\]

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

\[= 3 \cdot \left( a - \frac{2}{3} \right)(a + 1) =\]

\[= (3a - 2)(a + 1).\]

\[a^{2} - 9a - 10 = 0\]

\[a_{1} + a_{2} = 9;\ \ a_{1} \cdot a_{2} = - 10\]

\[a_{1} = 10,\ \ a_{2} = - 1.\]

\[a^{2} - 9a - 10 = (a - 10)(a + 1).\]

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

\[{(( - 1)}^{2} - 9 \cdot ( - 1) - 10) \cdot x =\]

\[= 3 \cdot ( - 1)² + ( - 1) - 2\]

\[0 \cdot x = 0\]

\[\Longrightarrow x - любое.\]

\[2)\ \ a = 10:\]

\[нет\ решения.\]

\[3)\ \ a = \frac{2}{3}:\]

\[x = \frac{3 \cdot \frac{2}{3} - 2}{\frac{2}{3} - 10} = \frac{2 - 2}{- 9\frac{1}{3}} =\]

\[= \frac{0}{- 9\frac{1}{3}} = 0.\]

\[4)\ \ a \neq - 1,\ a \neq 10,\ a \neq \frac{2}{3}:\]

\[x = \frac{3a - 2}{a - 10}.\]