Question

Решите подробно систему уравнений:
x^2 +y^2 = 27
xy=6

Answer 1

$left { {{x^2+y^2=27} atop {xy=6}} right.; left { {{x^2+2xy+y^2=27+2cdot 6} atop {y=6}} right. ; left { {{(x+y)^2=39} atop {xy=6}} right. ; left { {{x+y=pm sqrt{39}} atop {xy=6; ; ; }} right.\\1); ; x+y=sqrt{39}; ,; ; y=x-sqrt{39}; ,\\x(x-sqrt{39})=6; ,; ; x^2-xsqrt{39}-6=0; ,; ; D=39+24=63; ,\\x_1=frac{sqrt{39}-sqrt{63}}{2}=frac{sqrt3cdot (sqrt{13}-sqrt{21})}{2}; ,; ; x_2=frac{sqrt{39}+sqrt{63}}{2}=frac{sqrt3cdot (sqrt{13}+sqrt{21})}{2}$

$y_1=x_1-sqrt{39}=frac{sqrt{39}-sqrt{63}}{2}-sqrt{39}=-frac{sqrt{39}+sqrt{63}}{2}=-frac{sqrt3cdot (sqrt{13}+sqrt{21})}{2}\\y_2=x_2-sqrt{39}=frac{sqrt{39}+sqrt{63}}{2}-sqrt{39}=frac{-sqrt{39}+sqrt{63}}{2}=frac{sqrt3cdot (sqrt{21}-sqrt{13})}{2}\\2); ; x+y=-sqrt{39}; ; ,; ; y=-x-sqrt{39}\\x(-x-sqrt{39})=6; ; ,; ; x^2+xsqrt{39}+6=0; ,; ; D=39-24=15; ,\\x_1=frac{-sqrt{39}-sqrt{15}}{2}=-frac{sqrt3cdot (sqrt{13}+sqrt5)}{2}; ,; ; x_2=frac{-sqrt{39}+sqrt{15}}{2}=frac{sqrt3cdot (sqrt5-sqrt{13})}{2}$

$y_1=-x_1-sqrt{39}=frac{sqrt{39}+sqrt{15}}{2}-sqrt{39}=frac{-sqrt{39}+sqrt{15}}{2}=frac{sqrt3cdot (sqrt5-sqrt{13})}{2}\\y_2=-x_2-sqrt{39}=frac{sqrt{39}-sqrt{15}}{2}-sqrt{39}=frac{-sqrt{39}-sqrt{15}}{2}=-frac{sqrt3cdot (sqrt5+sqrt{13})}{2}\\Otvet:; ; Big (frac{sqrt3cdot (sqrt{13}-sqrt{21})}{2}, ;, -frac{sqrt3cdot (sqrt{13}+sqrt{21})}{2}Big ); ,; Big (frac{sqrt3cdot (sqrt{13}+sqrt{21})}{2}, ,, frac{sqrt3cdot (sqrt{21}-sqrt{13})}{2}Big ); ,$

$Big (-frac{sqrt3cdot (sqrt{5}+sqrt{13})}{2}, ,, frac{sqrt3cdot (sqrt5-sqrt{13})}{2}Big ); ,; Big (frac{sqrt3cdot (sqrt5-sqrt{13})}{2}, ,, -frac{sqrt3cdot (sqrt5+sqrt{13})}{2}Big ), .$

Answer 2

помоте пожалуйста https://znanija.com/task/31254790