a: \(\dfrac{2x^4-3x^3+4x^2+1}{x^2-1}=\dfrac{2x^4-2x^2-3x^3+3x+6x^2-6-3x+7}{x^2-1}\)
\(=2x^2-3x+6+\dfrac{-3x+7}{x^2-1}\)
Để dư bằng 0 thì -3x+7=0
=>x=7/3
b: \(\dfrac{x^5+2x^4+3x^2+x-3}{x^2+1}\)
\(=\dfrac{x^5+x^3+2x^4+2x^2-x^3-x+x^2+1+2x-4}{x^2+1}\)
\(=x^3+2x^2-x+1+\dfrac{2x-4}{x^2+1}\)
Để đư bằng 0 thì 2x-4=0
=>x=2