Khi thực hiện phép tính \(\left(\dfrac{1}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2-6x}{1-6x+9x^2}\), ta có kết quả là
\(\dfrac{18x^2-3x-1}{18x^2+6x}\).\(\dfrac{18x^2+3x+1}{18x^2+6x}\).\(\dfrac{6x^2+3x+1}{18x^2+6x}\).\(\dfrac{x^2+6x+9}{18x^2+6x}\).Hướng dẫn giải:\(\left(\dfrac{1}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2-6x}{1-6x+9x^2}\)
\(=\dfrac{3x+1+2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{6x\left(x-1\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{-6x^2+5x+1}{\left(1-3x\right)\left(1+3x\right)}.\dfrac{\left(1-3x\right)^2}{6x\left(x-1\right)}\)
\(=\dfrac{\left(1-x\right)\left(6x+1\right)}{\left(1-3x\right)\left(1+3x\right)}.\dfrac{\left(1-3x\right)^2}{6x\left(x-1\right)}\)
\(=\dfrac{-\left(6x+1\right)\left(1-3x\right)}{\left(1+3x\right)6x}\)
\(=\dfrac{-\left(-18x^2+3x+1\right)}{18x^2+6x}\)
\(=\dfrac{18x^2-3x-1}{18x^2+6x}\).
