a: \(\dfrac{3x^5-x^4-2x^3+x^2+4x+5}{x^2-2x+2}\)
\(=\dfrac{3x^5-6x^4+6x^3+5x^4-10x^3+10x^2+2x^3-4x^2+4x-5x^2+5}{x^2-2x+2}\)
\(=3x^3+5x^2+2x+\dfrac{-5x^2+5}{x^2-2x+2}\)
Để dư=0 thì \(-5x^2+5=0\)
=>\(x^2=1\)
=>\(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
b: \(\dfrac{2x^4-11x^3+19x^2-20x+9}{x^2-4x+1}\)
\(=\dfrac{2x^4-8x^3+2x^2-3x^3+12x^2-3x+5x^2-20x+5+3x+4}{x^2-4x+1}\)
\(=2x^2-3x+5+\dfrac{3x+4}{x^2-4x+1}\)
Dư=0
=>3x+4=0
=>x=-4/3
c: \(\dfrac{x^5+2x^4+3x^2+x-3}{x^2+1}\)
\(=\dfrac{x^5+x^3+2x^4+2x^2-x^3-x+x^2+1+2x-4}{x^2+1}\)
\(=x^3+2x^2-x+1+\dfrac{2x-4}{x^2+1}\)
Dư =0
=>2x-4=0
=>x=2
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