toán đại, giúp tui zới

$\begin{array}{l} {\left[ {f\left( {\sqrt {{x^2} + 9}  + 1} \right)} \right]^{2023}} + {\left[ {f\left( {{x^2} + x + 4} \right)} \right]^{2021}} = x + 2\left( * \right)\\ \left( * \right):x = 0 \Rightarrow f{\left( 4 \right)^{2023}} + f{\left( 4 \right)^{2021}} = 2\\ f\left( 4 \right) = u \Rightarrow {u^{2023}} + {u^{2021}} = 2\\ u\left( u \right) = {u^{2023}} + {u^{2021}},u \in \mathbb R \end{array}$

${D_{f\left( u \right)}} = \mathbb R,\forall {t_1},{t_2} \in \mathbb R,{t_1} \ne {t_2}$, giả sử $t_1<t_2$

$\begin{array}{l}
u = \dfrac{{f\left( {{t_1}} \right) - f\left( {{t_2}} \right)}}{{{t_1} - {t_2}}}\\
 = \dfrac{{{t_1}^{2023} + {t_1}^{2021} - {t_2}^{2023} - {t_2}^{2021}}}{{{t_1} - {t_2}}}\\
 = \dfrac{{\left( {{t_1} - {t_2}} \right).Q\left( u \right) + \left( {{t_1} - {t_2}} \right)P\left( u \right)}}{{{t_1} - {t_2}}}\left( {Q\left( u \right),P\left( u \right) > 0} \right)\\
 = Q\left( u \right) + P\left( u \right) > 0
\end{array}$

Suy ra hàm số $u$ đồng biến trên $\mathbb R$

$\begin{array}{l}
\forall {u_0} > 1 \Rightarrow {u^{2023}} + {u^{2021}} > {1^{2023}} + {1^{2021}} = 2\\
\forall {u_0} < 1 \Rightarrow {u^{2023}} + {u^{2021}} < {1^{2023}} + {1^{2021}} = 2\\
 \Rightarrow u = 1 \Rightarrow f\left( 4 \right) = 1
\end{array}$

Đạo hàm hai vế ta được:

$\begin{array}{l}
{\left[ {f\left( {\sqrt {{x^2} + 9}  + 1} \right)} \right]^{2023}} + {\left[ {f\left( {{x^2} + x + 4} \right)} \right]^{2021}} = x + 2\left( * \right)\\
\left( {{{\left[ {f\left( {\sqrt {{x^2} + 9}  + 1} \right)} \right]}^{2023}}} \right)'\left( 1 \right)\\
 = 2023\dfrac{x}{{\sqrt {{x^2} + 9} }}f'\left( {\sqrt {{x^2} + 9}  + 1} \right){\left[ {f\left( {\sqrt {{x^2} + 9}  + 1} \right)} \right]^{2022}}\\
\left( {{{\left[ {f\left( {{x^2} + x + 4} \right)} \right]}^{2021}}} \right)'\\
 = 2021\left( {2x + 1} \right)f'\left( {{x^2} + x + 4} \right){\left[ {f\left( {{x^2} + x + 4} \right)} \right]^{2020}}\left( 2 \right)\\
VT = \left( 1 \right) + \left( 2 \right) = 1\\
x = 0 \Rightarrow 2021f'\left( 4 \right)f{\left( 4 \right)^{2020}} = 1\\
 \Rightarrow 2021f'\left( 4 \right).1 = 1\\
 \Rightarrow f'\left( 4 \right) = \dfrac{1}{{2021}}\\
pttt:d:f'\left( 4 \right)\left( {x - 4} \right) + f\left( 4 \right)\\
\dfrac{1}{{2021}}\left( {x - 4} \right) + 1 = \dfrac{1}{{2021}}x + \dfrac{{2017}}{{2021}}
\end{array}$

$\begin{array}{l}
y = \dfrac{1}{{2021}}x + \dfrac{{2017}}{{2021}}\\
x = 0 \Rightarrow y = \dfrac{{2017}}{{2021}}\\
y = 0 \Rightarrow x =  - 2017\\
{S_{OAB}} = \dfrac{1}{2}\left| {xy} \right| = \dfrac{1}{2}\left| {\dfrac{{2017}}{{2021}}\left( { - 2017} \right)} \right|\\
 = \dfrac{{{{2017}^2}}}{{4042}}
\end{array}$