Bài 3:
a: \(=\dfrac{x^2+6x+9}{2\left(x+3\right)}=\dfrac{x+3}{2}\)
b: \(=\dfrac{\left(1-3x\right)\left(x-2\right)+2\left(x+2\right)-2-3x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-2-3x^2+6x+2x+4-2-3x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-3x^2+6x}{\left(x-2\right)\left(x+2\right)}=\dfrac{-3x}{x+2}\)

