Сократите дробь:a)x+4/x^2+7x+12б)5x^2+x-4/x^2+x

Решите задачу:

$\frac{x+4}{ x^{2} +7x+12} = \frac{x+4}{(x+4)(x+3)} = \frac{1}{x+3} \\ x^{2} +7x+12=0 \\ x_{1} + x_{2} =-7 \\ x_{1}* x_{2} =12 \\ x_{1} =-4 \\ x_{2} =-3$
$\frac{5 x^{2} +x-4}{ x^{2} +x} = \frac{5(x- \frac{4}{5})(x+1) }{x(x+1)}= \frac{(5x-4)(x+1)}{x(x+1)} = \frac{5x-4}{x} \\ 5 x^{2} +x-4=0 \\ D=1+4*4*5=81= 9^{2} \\ x_{1} = \frac{-1+9}{10} = \frac{8}{10} = \frac{4}{5} \\ x_{2} = \frac{-1-9}{10} =-1$