Bài 2:
\(B=\left[\left(\dfrac{x+1-3}{x-2}-3x\right)\cdot\dfrac{x-2}{1-3x}\right]-\dfrac{x^2+4}{x-2}\)
\(=\left(\dfrac{x-2}{x-2}-3x\right)\cdot\dfrac{x-2}{1-3x}-\dfrac{x^2+4}{x-2}\)
\(=x-2-\dfrac{x^2+4}{x-2}=\dfrac{x^2-4x+4-x^2-4}{x-2}=\dfrac{-4x}{x-2}\)