Система срочно x+y=xy x^2+y^2=4xy

Question

Система срочно
x+y=xy
x^2+y^2=4xy

Answer 1

X^2+2xy+y^2=6xy
(x+y)^2=6xy
(xy)^2=6xy
xy(xy-6)=0
1) xy=0
x+y=xy
x+y=0
xy=0
x=0
y=0
2) xy=6
x+y=xy
x+y=6
x=6-y
(6-y)y=6
y^2-6y+6=0
D=12
y1=(6+2sqrt(3))/2=3+sqrt(3)
y2=3-sqrt(3)
x1=6-3-sqrt(3)=3-sqrt(3)
x2=6-3+sqrt(3)=3+sqrt(3)
Ответ: (0;0), (3-sqrt(3);3+sqrt(3)), (3+sqrt(3);3-sqrt(3))

Answer 2

{x+y = xy
{x² + y² = 4xy

{x+y = xy
{x² + y² + 2xy = 6xy

{x+y = xy
{(x+y)² = 6xy

{x+y=xy
{x²y² = 6xy

{x+y = xy
{xy(xy-6) = 0

{x+y = xy
x = 0 или y = 0 или xy= 6
1) x = 0 => y = 0 - первый ответ
2) xy = 6

{x+y = 6
{x² + y² = 24

{y = 6-x
{x² + 36 - 12x + x² = 24

{y = 6-x
{2x² - 12x + 12 = 0

x² - 6x + 6 = 0
D = 36 - 24 = 12
x1 = (6-2√3)/2 = 3-√3 => y1 = 3 + √3
x2 = 3+√3 => y2 = 3-√3

Ответ: (0; 0); (3-√3; 3+√3); (3+√3; 3-√3)