Question

Answer 1

Bài 1:

\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}=x-2\)

Bài 2:

a: \(=\dfrac{3x^2-x+3-x^2+2x-1-2x^2-2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{-x}{\left(x-1\right)\left(x^2+x+1\right)}\)

b: \(=\dfrac{-x\left(2y+x\right)+x\left(2y-x\right)+4xy}{\left(2y-x\right)\left(2y+x\right)}\)

\(=\dfrac{-2xy-x^2+2xy-x^2+4xy}{\left(2y-x\right)\left(2y+x\right)}=\dfrac{-2x^2+4xy}{\left(2y-x\right)\left(2y+x\right)}\)

\(=\dfrac{-2x\left(x-2y\right)}{-\left(x-2y\right)\left(x+2y\right)}=\dfrac{2x}{x+2y}\)