Желательно на бумаге

Решите задачу:

$\frac{(a ^{2} -a-12)(a ^{2} -1)}{(a+3)(a^{2} -5a+4)} \\ a ^{2} -a-12=0 \\ D=1+48=49 \\ \sqrt{D} =7 \\ a _{1} = \frac{1+7}{2} = \frac{8}{2} =4 \\ a_{2} = \frac{1-7}{2} = \frac{-6}{2} =-3 \\ a^{2} -a-12=(a-4)(a+3) \\ a^{2} -5a+4=0 \\ D=25-16=9 \\ \sqrt{D} =3 \\ a _{1} = \frac{5+3}{2} = \frac{8}{2} =4 \\ a _{2} = \frac{5-3}{2} = \frac{2}{2} =1 \\ a x^{2} -5a+4=(a-4)(a-1) \\ \\$

$\frac{(a ^{2} -a-12)(a ^{2} -1)}{(a+3)(a^{2} -5a+4)} = \frac{(a-4)(a+3)(a-1)(a+1)}{(a+3)(a-4)(a-1)} =a+1$

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