If x + y = 3, xy = 1, then x2 + Y2=______ ．

If x + y = 3, xy = 1, then x2 + Y2=______ ．


x2+y2=x2+2xy+y2-2xy，=（x+y）2-2xy，=9-2，=7．



Given | (x + y) 2-4 | + [(X-Y) 2-3] 2 + 0, find the values of x2 + Y2 and XY


|(x+y)2-4|+[(x-y)2-3]2=0
=>(x+y)^2-4 =0 and (x-y)^2-3 =0
=>(x+y)^2 = 4 (1) and
(x-y)^2 = 3 (2)
(1) -(2)
4xy =1
xy =1/4
(1)+(2)
2(x^2+y^2) =7
x^2+y^2 = 7/2



If x2 + y2 = 1, then (XY) / (x + y)


Well, 2 * (x + y) = 1, so x + y = half
Square of (x2 + Y2) = 1
It's square root deformation, and then it's solved
That's the general idea

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