Sau khi rút gọn phân thức \(\dfrac{3x+18}{x^2-4}.\dfrac{x^2+4x+4}{x^2+8x+12}\), ta có kết quả là
\(\dfrac{3}{x-2}\).\(\dfrac{3}{x+2}\).\(\dfrac{-3}{x-2}\).\(\dfrac{3}{\left(x-2\right)\left(x+2\right)}\).Hướng dẫn giải:\(\dfrac{3x+18}{x^2-4}.\dfrac{x^2+4x+4}{x^2+8x+12}\)
\(=\dfrac{3\left(x+6\right)}{\left(x-2\right)\left(x+2\right)}.\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x+6\right)}\)
\(=\dfrac{3}{x-2}\).