If | x + Y-5 | + (xy-6) ^ 2 = 0, then x ^ 2 + y ^ 2=

If | x + Y-5 | + (xy-6) ^ 2 = 0, then x ^ 2 + y ^ 2=

thirteen

How to solve the system of equations x + y = 9, xy = 12

From x + y = 9, we can get that:
x=9-y
Substituting x = 9-y into xy = 12, we get:
(9-y)y=12
y^2-9y+12=0
The solution is as follows
y(1,2)=(9±√33)/2
So,
x=(9+√33)/2,y=(9-√33)/2
or
x=(9-√33)/2,y=(9+√33)/2

3 / (2 + x)] + [3 / (2 + y)] = 1, then the minimum value of XY is,,, to process, thank you!

3／（2+x）+3／（2+y）=1
General division, to denominator 3Y + 6 + 3x + 6 = XY + 2x + 2Y + 4
Xy = x + y + 8 > = 2 radical sign XY + 8 commutation so that radical sign xy = t
T ^ 2-2t-8 > = 0
(t-4)(t+2)>=0
t>=4
xy>=16
So the minimum value of XY is 16
Or 3 / (2 + x) + 3 / (2 + y) = 1
Let 3 / (2 + x) = sin ^ 2a, 3 / (2 + y) = cos ^ 2A
x=(3-2sin^2a)/sin^2a,
y=(3-2cos^2a)/cos^2a.
There are
XY=[(3-2sin^2a)/sin^2a]*[(3-2cos^2a)/cos^2a]
=[3+4(sina*cosa)^2]/[(sina*cosa)^2]
={3/[(sina*cosa)^2]}+4
=[12/sin(2a)]+4.
For XY to have a minimum, sin (2a) must be the largest, and sin (2a) max = 1
There are,
XY min = 12 + 4 = 16