Question

Rút gọn:

\(\left[\dfrac{2a^3+a^2-a}{a^3-1}-2+\dfrac{1}{1-a}\right]:\left[1:\dfrac{2a-1}{a-a^2}\right]\)

Answer 1

\(\left(\dfrac{2a^3+a^2-a}{a^3-1}-2+\dfrac{1}{1-a}\right):\left(1:\dfrac{2a-1}{a-a^2}\right)\)

\(=\left(\dfrac{2a^3+a^2-a-2a^3+2-a^2-a-1}{\left(a-1\right)\left(a^2+a+1\right)}\right):\left(\dfrac{a\left(1-a\right)}{2a-1}\right)\)

\(=\dfrac{-2a+1}{\left(a-1\right)\left(a^2+a+1\right)}.\dfrac{2a-1}{a\left(1-a\right)}\)

\(=\dfrac{6a-3}{\left(a-1\right)^2\left(a^2+a+1\right)}\)

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Thấy sai sai :vv