\(f\left(x\right)=ax^2+bx+2020\\ \Leftrightarrow f\left(\sqrt{3}-1\right)=a\left(4-2\sqrt{3}\right)+b\left(\sqrt{3}-1\right)+2020=2021\\ \Leftrightarrow4a-2a\sqrt{3}+b\sqrt{3}-b-1=0\\ \Leftrightarrow\left(4a-b-1\right)-\sqrt{3}\left(2a-b\right)=0\\ \Leftrightarrow4a-b-1=\sqrt{3}\left(2a-b\right)\)
Vì a,b hữu tỉ nên \(4a-b-1;2a-b\) hữu tỉ
Mà \(\sqrt{3}\) vô tỉ nên \(\sqrt{3}\left(2a-b\right)\) hữu tỉ khi \(2a-b=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}4a-b-1=0\\2a-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\end{matrix}\right.\)
\(\Leftrightarrow f\left(1+\sqrt{3}\right)=\dfrac{1}{2}\left(4+2\sqrt{3}\right)+1+\sqrt{3}+2020=2023+2\sqrt{3}\)
