Ta có `:`

`A ( x ) = 12x^3 + 2ax + a^2`

`=> A ( 1 ) = 12 . 1^3 + 2 . a . 1 + a^2`

`=> A ( 1 ) = 12 + 2a + a^2`

`B ( x ) = 2x^2 - | 2a + 3 |x + a^2`

`=> B ( -2 ) = 2 . ( -2 )^2 - | 2a + 3 | . ( -2 )+ a^2`

`=> B ( -2 ) = 8 - | 2a + 3 | . ( -2 ) + a^2`

Vì `A ( 1 ) = B ( -2 )`

`=> 12 + 2a + a^2 = 8 - | 2a + 3 | . ( -2 ) + a^2`

`=> 12 - 8 + 2a + a^2 - a^2 = - | 2a + 3 | . ( -2 )`

`=> 4 + 2a = | 2a + 3 | . 2`

`=> 2 + a = | 2a + 3 |`

`=> 2 + a = 2a + 3` hoặc `2 + a = -2a - 3`

`=> a - 2a = 3 - 2` hoặc `a + 2a = -3 - 2`

`=> -a = 1` hoặc `3a = -5`

`=> a = -1` hoặc `a = -5/3`

Vậy `:` `a in { -1 ; -5/3 }`

`A(x) = 12x^3 + 2ax + a^2`

`=> A(1) = 12 .1^3 + 2a.1 + a^2`

`=> A(1) = 12 + 2a + a^2`

`B(x) = 2x^2 - |2a + 3|x + a^2`

`=> B(-2) = 2 .(-2)^2 - |2a + 3|.(-2) + a^2`

`=> B(-2) = 8 - |2a + 3|.(-2) + a^2`

Mà `A(1) = B(-2)`

`=> 12 + 2a + a^2 = 8 - |2a + 3|.(-2)`

`=> 12 + 2a = 8 - |2a + 3|.(-2)`

`=> 12 + 2a - 8 = -|2a + 3|.(-2)`

`=> 4 + 2a = |2a + 3|.2`

`=>(4 + 2a) : 2 = |2a + 3|`

`=> 2 + a = |2a +3|`

`=>` $\left[\begin{matrix} 2 + a = 2a + 3\\ 2 + a = 2a - 3\end{matrix}\right.$

`=>` $\left[\begin{matrix} a - 2a = 3 -2\\ a + 2a = -3 - 2\end{matrix}\right.$

`=>` $\left[\begin{matrix} -a = 1\\ 3a = -5\end{matrix}\right.$

`=>` $\left[\begin{matrix} a=-1\\ a=\dfrac{-5}{3}\end{matrix}\right.$

Vậy `a= {-1; -5/3}` 

`#HPHG`