Khi thực hiện phép tính \(\dfrac{4x^2+3}{3x\left(3x+1\right)}+\dfrac{-2x+1}{3x+1}\), ta có kết quả là
\(\dfrac{-2x^2+3x+3}{3x\left(3x+1\right)}\).\(\dfrac{-2x^2-3x+3}{3x\left(3x+1\right)}\).\(\dfrac{1}{3x+1}\).\(\dfrac{1}{3}\).Hướng dẫn giải:\(\dfrac{4x^2+3}{3x\left(3x+1\right)}+\dfrac{-2x+1}{3x+1}\)
\(\dfrac{4x^2+3}{3x\left(3x+1\right)}+\dfrac{\left(-2x+1\right).3x}{3x\left(3x+1\right)}\)
\(=\dfrac{4x^2+3-6x^2+3x}{3x\left(3x+1\right)}\)
\(=\dfrac{-2x^2+3x+3}{3x\left(3x+1\right)}\).