ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};\pm3\right\}\).
Ta viết lại được biểu thức thành:
\(A=\left[\dfrac{\left(x-1\right)\left(3-x\right)}{\left(3-x\right)\left(3+x\right)}-\dfrac{2\left(3+x\right)}{\left(3-x\right)\left(3+x\right)}+\dfrac{x^2+3}{\left(3-x\right)\left(3+x\right)}\right]:\dfrac{-2}{2x+1}\)
\(=\dfrac{\left(x-1\right)\left(3-x\right)-2\left(3+x\right)+x^2+3}{\left(3-x\right)\left(3+x\right)}:\dfrac{-2}{2x+1}\)
\(=\dfrac{-x^2+4x-3-6-2x+x^2+3}{\left(3-x\right)\left(3+x\right)}:\dfrac{-2}{2x+1}\)
\(=\dfrac{2x-6}{\left(3-x\right)\left(3+x\right)}:\dfrac{-2}{2x+1}\)
\(=\dfrac{-2\left(3-x\right)}{\left(3-x\right)\left(3+x\right)}\cdot\dfrac{2x+1}{-2}\)
\(=\dfrac{2x+1}{x+3}\).
Vậy: \(A=\dfrac{2x+1}{x+3}\)
