Question

желательно на бумаге

Answer 1

$a^2-a-12=0 \ D=49 \a_1=4 , a_2=-3$

$a^2-5a+4=0 \ D=9 \ a_1=4 , a_2=1$

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

Answer 2

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

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