решите систему уравнений

А) (x/x+3) + (x+3/x) = 17/4

x^2-4xy+4y^2=0

Б) (y+1/y+2) + (y+2/y+1) = 25/12

x^2-2xy+y^2=0

1

{x²-4xy+4y²=0⇒ (x-2y) ²=0⇒x-2y=0⇒x=2y

{x / (x+3) + (x+3) / x=17/4⇒2y / (2y+3) + (2y+3) / 2y=17/4

(2y+3) / 2y=a

1/a+a=17/4

4a²-17a+4=0

D=289-64=225

a1 = (17-15) / 8=1/4⇒ (2y+3) / 2y=1/4

8y+12=2y

6y=-12

y1=-2⇒x1=-4

a2 = (17+15) / 8=4⇒ (2y+3) / 2y=4

2y+3=8y

6y=3

y2=0,5⇒x2=1

(-4; -2) ; (1; 0,5)

2

{x²-2xy+y²=0⇒ (x-y) ²=0⇒x-y=0⇒x=y

{ (y+1) / (y+2) + (y+2) / (y+1) = 25/12

(y+2) / (y+1) = a

a+1/a=25/12

12a²-25a+12=0

D=625-576=49

a1 = (25-7) / 24=3/4⇒ (y+2) / (y+1) = 3/4

4y+8=3y+3

y1=-5⇒x1=-5

a2 = (25+7) / 24=4/3⇒ (y+2) / (y+1) = 4/3

3y+6=4y+4

y2=2⇒x2=2

(-5; -5) ; (2; 2)