Question

(x+3)²+y^2+(x-y)^2有实根,求xy

若x+y=4,求(x^2+y^2)/2+xy的值.
解方程组{（x+3)^2-(y-2)^2=(x+y)(x-y)
{3x-y=2

Answer 1

(x²+y²)/2 +xy=(x²+y²+2xy)/2=(x+y)²/2=4²/2=8
(x+3)²-(y-2)²=(x+y)(x-y) (1)
3x-y=2 (2)
由(1)得x²+6x+9-y²+4y-4=x²-y²
6x+4y=-5 (3)
(3)+(2)×4
18x=3 x=1/6,代入(2)
y=3x-2=3×(1/6)-2=1/2-2=-3/2
x=1/6 y=-3/2