Question

3a в квадрате +4а+1 / (a+1) в квадрате - a-1/a+1

Answer 1

$\frac{3a^2+4a+1}{(a+1)^2} - \frac{a-1}{a+1} = \frac{3a^2+4a+1-(a-1)(a+1)}{(a+1)^2} = \frac{3a^2+4a+1-a^2+1}{(a+1)^2} = \frac{2(a+1)^2}{(a+1)^2} = 2$