\(x^{20}+x^{11}-x^{2004}=\left[\left(x^2\right)^{10}-1\right]+x\left[\left(x^2\right)^5-1\right]-\left[\left(x^2\right)^{1002}-1\right]+x\)
\(=\left(x^2-1\right)A\left(x\right)+x\left(x^2-1\right)B\left(x\right)-\left(x^2-1\right)C\left(x\right)+x\)
Vậy số dư là: x