Question

сделайте пожалуйста б) и номер 2​

Answer 1

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

$\left \{ {{5x-3y=-6}|*-2 \atop {2x-5y=10}|*5} \right.\\\\+\left \{ {{-10x+6y=12} \atop {10x-25y=50}} \right.\\-------\\-19y=62\\y=-3\frac{5}{19}\\\\2x=10+5y=10+5*(-3\frac{5}{19})=10-16\frac{6}{19}=-6\frac{6}{19}\\\\Otvet:\boxed{(-6\frac{6}{19};-3\frac{5}{19})}$