\(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3x}-x-1\right)\right]:\dfrac{x-1}{x}\\ =\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3x}-\left(x+1\right)\right)\right].\dfrac{x}{x-1}\\ =\left[\dfrac{2}{3x}-\dfrac{2}{x+1}.\left(x+1\right)\left(\dfrac{1}{3x}-1\right)\right].\dfrac{x}{x-1}\\ =\left[\dfrac{2}{3x}-2\left(\dfrac{1}{3x}-1\right)\right].\dfrac{x}{x-1}\\ =\left[2\left(\dfrac{1}{3x}-\dfrac{1}{3x}+1\right)\right].\dfrac{x}{x-1}\\ =\dfrac{2x}{x-1}\)
Đúng 0
Bình luận (1)
