Question

ПОМОГИТЕ ПЛИЗЗЗ!!! решите систему
ху=6
у^2 - 4х=1

Answer 1

$left { {{xy=6 => y= frac{6}{x} } atop {y^2-4x=1}} right. => \ (frac{6}{x})^2-4x=1 \ frac{36}{x^2}-4x=1|*x^2; x neq 0 \ 36-4x^3= x^{2} \ x^{2} +4x^3-36=0 \ (x-2)(4 x^{2} +9x+18)=0 =>\ 4 x^{2} +9x+18 neq 0; (D=81-4*4*18<0) \ x-2=0 => x=2 \ y=frac{6}{2}=3 \ (2;3)$