Đáp án:
\[\left[ \begin{array}{l}
x = 2\\
x = 2 \pm \sqrt 2
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) = {x^2} - 2x\\
\Rightarrow f'\left( x \right) = 2x - 2\\
\Rightarrow f'\left( x \right) = 0 \Leftrightarrow x = 1\\
y = f\left( {f\left( { - x + 3} \right)} \right)\\
\Rightarrow y' = \left( {f\left( { - x + 3} \right)} \right)'.f'\left( {f\left( { - x + 3} \right)} \right)\\
\Leftrightarrow y' = \left( { - x + 3} \right)'.f'\left( { - x + 3} \right).f'\left( {f\left( { - x + 3} \right)} \right)\\
\Leftrightarrow y' = - f'\left( { - x + 3} \right).f'\left( {f\left( { - x + 3} \right)} \right)\\
y' = 0 \Leftrightarrow \left[ \begin{array}{l}
f'\left( { - x + 3} \right) = 0\\
f'\left( {f\left( { - x + 3} \right)} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- x + 3 = 1\\
f\left( { - x + 3} \right) = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
{\left( { - x + 3} \right)^2} - 2.\left( { - x + 3} \right) = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
{x^2} - 6x + 9 + 2x - 6 = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
{x^2} - 4x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 2 \pm \sqrt 2
\end{array} \right.
\end{array}\)
