Question

скоротіть дріб (а^2-1)/(4а^2+а-3)

Answer 1

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