Решение: a 2 + a x − 2 x 2 − 6 a − 3 x + 9 ∣ x ∣ = 0 a^{2} + ax - 2 x^{2} - 6 a - 3 x + 9 \left|x\right| = 0 a 2 + a x − 2 x 2 − 6 a − 3 x + 9 ∣ x ∣ = 0 [ { x ≤ 0 a 2 + a x − 2 x 2 − 6 a − 3 x − 9 x = 0 { x > 0 a 2 + a x − 2 x 2 − 6 a − 3 x + 9 x = 0 \left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}x \leq 0 \\ a^{2} + ax - 2 x^{2} - 6 a - 3 x - 9 x = 0\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}x > 0 \\ a^{2} + ax - 2 x^{2} - 6 a - 3 x + 9 x = 0\end{matrix} \end{matrix}\right. \end{matrix}\right. { x ≤ 0 a 2 + a x − 2 x 2 − 6 a − 3 x − 9 x = 0 { x > 0 a 2 + a x − 2 x 2 − 6 a − 3 x + 9 x = 0 [ { x ≤ 0 a 2 + a x − 2 x 2 − 6 a − 12 x = 0 ( 1 ) { x > 0 a 2 + a x − 2 x 2 − 6 a + 6 x = 0 ( 2 ) \left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}x \leq 0 \\ a^{2} + ax - 2 x^{2} - 6 a - 12 x = 0\end{matrix} \end{matrix}\right ( 1 ) \\ \left\{\begin{matrix} \begin{matrix}x > 0 \\ a^{2} + ax - 2 x^{2} - 6 a + 6 x = 0\end{matrix} \end{matrix}\right ( 2 ) \end{matrix}\right. { x ≤ 0 a 2 + a x − 2 x 2 − 6 a − 12 x = 0 ( 1 ) { x > 0 a 2 + a x − 2 x 2 − 6 a + 6 x = 0 ( 2 ) ( 1 ) a 2 + a x − 6 a − 12 x − 2 x 2 = 0 ; D > 0 \left(1\right) a^{2} + ax - 6 a - 12 x - 2 x^{2} = 0 ; D > 0 ( 1 ) a 2 + a x − 6 a − 12 x − 2 x 2 = 0 ; D > 0 a 2 + ( x − 6 ) a − 12 x − 2 x 2 = 0 a^{2} + \left(x - 6\right) a - 12 x - 2 x^{2} = 0 a 2 + ( x − 6 ) a − 12 x − 2 x 2 = 0 D = ( x − 6 ) 2 − 4 ⋅ ( − 12 x − 2 x 2 ) D = \left(x - 6\right)^{2} - 4 \cdot \left(- 12 x - 2 x^{2}\right) D = ( x − 6 ) 2 − 4 ⋅ ( − 12 x − 2 x 2 ) = x 2 − 12 x + 36 + 48 x + 8 x 2 = = x^{2} - 12 x + 36 + 48 x + 8 x^{2} = = x 2 − 12 x + 36 + 48 x + 8 x 2 = 9 x 2 + 36 x + 36 = ( 3 x + 6 ) 2 9 x^{2} + 36 x + 36 = \left(3 x + 6\right)^{2} 9 x 2 + 36 x + 36 = ( 3 x + 6 ) 2 a 1 = 6 − x + 3 x + 6 2 = 12 + 2 x 2 = 6 + x a_{1} = \frac{6 - x + 3 x + 6}{2} = \frac{12 + 2 x}{2} = 6 + x a 1 = 2 6 − x + 3 x + 6 = 2 12 + 2 x = 6 + x a 2 = 6 − x − 3 x + 6 2 = − 4 x 2 = − 2 x . a_{2} = \frac{\cancel{6} - x - 3 x + \cancel{6}}{2} = \frac{- 4 x}{2} = - 2 x . a 2 = 2 6 − x − 3 x + 6 = 2 − 4 x = − 2 x . ( 2 ) a 2 + a x − 2 x 2 − 6 a + 6 x = 0 ; D > 0 \left(2\right) a^{2} + ax - 2 x^{2} - 6 a + 6 x = 0 ; D > 0 ( 2 ) a 2 + a x − 2 x 2 − 6 a + 6 x = 0 ; D > 0 a 2 + a x − 6 a + 6 x − 2 x 2 = 0 a^{2} + ax - 6 a + 6 x - 2 x^{2} = 0 a 2 + a x − 6 a + 6 x − 2 x 2 = 0 a 2 + ( x − 6 ) a + 6 x − 2 x 2 = 0 a^{2} + \left(x - 6\right) a + 6 x - 2 x^{2} = 0 a 2 + ( x − 6 ) a + 6 x − 2 x 2 = 0 D = ( x − 6 ) 2 − 4 ⋅ ( 6 x − 2 x 2 ) = D = \left(x - 6\right)^{2} - 4 \cdot \left(6 x - 2 x^{2}\right) = D = ( x − 6 ) 2 − 4 ⋅ ( 6 x − 2 x 2 ) = x 2 − 12 x + 36 − 24 x + 8 x 2 = x^{2} - 12 x + 36 - 24 x + 8 x^{2} = x 2 − 12 x + 36 − 24 x + 8 x 2 = 9 x 2 − 36 x + 36 = ( 3 x − 6 ) 2 9 x^{2} - 36 x + 36 = \left(3 x - 6\right)^{2} 9 x 2 − 36 x + 36 = ( 3 x − 6 ) 2 a 1 = 6 − x + 3 x − 6 2 = 2 x 2 = x a_{1} = \frac{6 - x + 3 x - 6}{2} = \frac{2 x}{2} = x a 1 = 2 6 − x + 3 x − 6 = 2 2 x = x a 2 = 6 − x − 3 x + 6 2 = 12 − 4 x 2 = 6 − 2 x a_{2} = \frac{6 - x - 3 x + 6}{2} = \frac{12 - 4 x}{2} = 6 - 2 x a 2 = 2 6 − x − 3 x + 6 = 2 12 − 4 x = 6 − 2 x [ { x ≤ 0 [ a = 6 + x = − 2 x { x > 0 [ a = x a = 6 − 2 x \left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}x \leq 0 \\ \left[\begin{matrix} \begin{matrix}a = 6 + x \\ = - 2 x\end{matrix} \end{matrix}\right.\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}x > 0 \\ \left[\begin{matrix} \begin{matrix}a = x \\ a = 6 - 2 x\end{matrix} \end{matrix}\right.\end{matrix} \end{matrix}\right. \end{matrix}\right. ⎩ ⎨ ⎧ x ≤ 0 [ a = 6 + x = − 2 x ⎩ ⎨ ⎧ x > 0 [ a = x a = 6 − 2 x a = 6 − 2 x a = 6 - 2 x a = 6 − 2 x
Ответ: ( − ∞ ; 0 ] ∩ { 2 } ∪ [ 6 ; + ∞ ) ( - \infty ; 0 \left]\right. \cap \left\{2\right\} \cup [ 6 ; + \infty ) ( − ∞ ; 0 ] ∩ { 2 } ∪ [ 6 ; + ∞ )
Источник : ФИПИ
