Ta có: \(\left(a^2-a+1\right)\left(a+2\right)\)
\(=a^3+2a^2-a^2-2a+a+2\)
\(=a^3+a^2-a+2\)(1)
Ta có: \(\left(a^2+3a-1\right)\left(a-2\right)\)
\(=a^3-2a^2+3a^2-6a-a+2\)
\(=a^3+a^2-7a+2\)(2)
Từ (1) và (2) suy ra \(\dfrac{a^2-a+1}{a-2}\ne\dfrac{a^2+3a-1}{a+2}\)