Cho \(x=0\Rightarrow f^3\left(2\right)-2f^2\left(2\right)=0\Leftrightarrow\left[{}\begin{matrix}f\left(2\right)=0\\f\left(2\right)=2\end{matrix}\right.\)
Đạo hàm 2 vế giả thiết:
\(-3f^2\left(2-x\right).f'\left(2-x\right)-12f\left(2+3x\right).f'\left(2+3x\right)+2x.g\left(x\right)+x^2.g'\left(x\right)+36=0\)
Cho \(x=0\Rightarrow-3f^2\left(2\right).f'\left(2\right)-12f\left(2\right).f'\left(2\right)+36=0\) (1)
- Với \(f\left(2\right)=0\) thay vào \(\left(1\right)\Leftrightarrow36=0\) (vô lý)
\(\Rightarrow f\left(2\right)=2\)
Thay vào (1) \(\Rightarrow-12.f'\left(2\right)-24f'\left(2\right)+36=0\)
\(\Rightarrow f'\left(2\right)=1\)
\(\Rightarrow A=3.2+4.1\)
