Question

Решить Систему
xy=10y+20
x(y+2)=12(y+5)

Answer 1

$begin{cases}xy=10y+20\x(y+2)=12(y+5)end{cases} textless = textgreater begin{cases}xy=10y+20\x(y+2)=12(y+5)end{cases}\x=frac{12(y+5)}{y+2}\\frac{12y(y+5)}{y+2}=10y+20\12y^2+60y=(10y+20)(y+2)\12y^2+60y=10y^2+20y+20y+40\2y^2+20y-40=0\y^2+10y-20=0\y_1=-5+3sqrt{5} y_2=-5-3sqrt{5}$
$x_1=frac{12(-5+3sqrt5+5)}{(-5+3sqrt5+2)}=frac{36sqrt5(3sqrt5+3)}{(-3+3sqrt5)(3sqrt5+3)}=frac{540+108sqrt5}{36}=15+3sqrt5\x_2=frac{12(-5-3sqrt5+5)}{(-5-3sqrt5+2)}=frac{-36sqrt5(3sqrt5-3)}{-(3+3sqrt5)(3sqrt5-3)}=-frac{-540+108sqrt5}{36}=15-3sqrt5\OTBET:(15+3sqrt5;-5+3sqrt5);(15-3sqrt5;-5-3sqrt5)$