Khi rút gọn biểu thức \(\dfrac{4a^2-3a+5}{a^3-1}-\dfrac{1-2a}{a^2+a+1}-\dfrac{6}{a-1}\), ta có kết quả là
\(\dfrac{-12a}{a^3-1}\).\(\dfrac{9a}{a^3-1}\).\(\dfrac{9}{a^2+a+1}\).\(\dfrac{9a}{a^2+a+1}\).Hướng dẫn giải:\(\dfrac{4a^2-3a+5}{a^3-1}-\dfrac{1-2a}{a^2+a+1}-\dfrac{6}{a-1}\)
\(=\dfrac{4a^2-3a+5-\left(1-2a\right)\left(a-1\right)-6\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{4a^2-3a+5-\left(-2a^2+3a-1\right)-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-12a}{a^3-1}\).
