\(u=-x^2+2x=1-\left(x-1\right)^2\le1\)
\(y=1+\dfrac{2024}{f\left(u\right)}\Rightarrow y'=-\dfrac{2024.u'.f'\left(u\right)}{\left[f\left(u\right)\right]^2}=0\)
\(\Rightarrow\left[{}\begin{matrix}u'=0\Rightarrow x=1\\f'\left(u\right)=0\end{matrix}\right.\)
\(f'\left(u\right)=0\Rightarrow u=\left\{-1;0;1\right\}\)
\(\Rightarrow-x^2+2x=\left\{-1;0;1\right\}\)
Hàm có 5 cực trị mới đúng