Đáp án:
\[\left\{ \begin{array}{l}
a = - 2\\
b = 1\\
c = - 1
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) = F'\left( x \right)\\
\Leftrightarrow \left( {2{x^2} - 5x + 2} \right).{e^{ - x}} = \left[ {\left( {a{x^2} + bx + c} \right).{e^{ - x}}} \right]'\\
\Leftrightarrow \left( {2{x^2} - 5x + 2} \right).{e^{ - x}} = \left( {a{x^2} + bx + c} \right)'.{e^{ - x}} + \left( {a{x^2} + bx + c} \right).\left( {{e^{ - x}}} \right)'\\
\Leftrightarrow \left( {2{x^2} - 5x + 2} \right).{e^{ - x}} = \left( {2ax + b} \right).{e^{ - x}} + \left( {a{x^2} + bx + c} \right).\left( { - {e^{ - x}}} \right)\\
\Leftrightarrow \left( {2{x^2} - 5x + 2} \right).{e^{ - x}} = \left( { - a{x^2} + \left( {2a - b} \right)x + \left( {b - c} \right)} \right){e^{ - x}}\\
\Leftrightarrow \left\{ \begin{array}{l}
- a = 2\\
2a - b = - 5\\
b - c = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = - 2\\
b = 1\\
c = - 1
\end{array} \right.
\end{array}\)
