Đáp án:
\[\left[ \begin{array}{l}
x = 1\\
x = \frac{2}{3}\\
x = \frac{4}{3}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) = {x^2} - 2x\\
\Rightarrow f'\left( x \right) = 2x - 2\\
\Rightarrow f'\left( x \right) = 0 \Leftrightarrow x = 1\\
y = {f^4}\left( {3x - 2} \right)\\
\Rightarrow y' = 4.\left[ {f\left( {3x - 2} \right)} \right]'.{f^3}\left( {3x - 2} \right)\\
\Leftrightarrow y' = 4.\left( {3x - 2} \right)'.f'\left( {3x - 2} \right).{f^3}\left( {3x - 2} \right)\\
\Leftrightarrow y' = 12f'\left( {3x - 2} \right).{f^3}\left( {3x - 2} \right)\\
y' = 0 \Leftrightarrow \left[ \begin{array}{l}
f'\left( {3x - 2} \right) = 0\\
f\left( {3x - 2} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 2 = 1\\
{\left( {3x - 2} \right)^2} - 2.\left( {3x - 2} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 2 = 1\\
3x - 2 = 0\\
3x - 2 = 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{2}{3}\\
x = \frac{4}{3}
\end{array} \right.
\end{array}\)
Vậy \(\left[ \begin{array}{l}
x = 1\\
x = \frac{2}{3}\\
x = \frac{4}{3}
\end{array} \right.\)
