Question

скоротіть дріб (а^2-1)/(4а^2+а-3)

Answer 1

$frac{a^2-1}{4a^2+a-3}=frac{(a-1)(a+1)}{4a^2+a-3}\\4a^2+a-3=0\D^2=1-4*4(-3)=1+16*3=49; D=+-7\a_1=frac{-1+7}{6}=1\a_2=frac{-1-7}{6}=frac{-8}{6}=frac{-4}{3}\frac{a^2-1}{4a^2+a-3}=frac{(a-1)(a+1)}{4a^2+a-3}=frac{a^2-1}{4a^2+a-3}=frac{(a-1)(a+1)}{4(a+frac{4}{3})(a-1))}=frac{a+1}{4(a+frac{4}{3})}=frac{a+1}{4a+frac{16}{3}}=frac{3(a+1)}{12a+16}=frac{3(a+1)}{4(3a+4)}$