Hoặc
b) a+3a2−1−1a2+a;
b) a+3a2−1−1a2+a
=a+3(a+1)(a−1)−1a(a+1)
=a(a+3)a(a+1)(a−1)−a−1a(a+1)(a−1)
=a2+3a−a+1a(a+1)(a−1)
=a2+2a+1a(a+1)(a−1)
=(a+1)2a(a+1)(a−1)=a+1a(a−1).