Question

解方程组 x2+2xy+y2-4=0 x2-2xy+y2-1=0

Answer 1

（x+y）²=2²
(x-y)²=1
∴x+y=±4
x-y=±1
(1)x+y=2
x-y=1
那么x=3/2 y=1/2
(2)x+y=2
x-y=-1
那么x=1/2 y=3/2
(2)x+y=-2
x-y=1
那么x=-1/2 y=-3/2
(4)x+y=-2
x-y=-1
那么x=-3/2 y=-1/2

Answer 2

x=1.5 y=0.5
x=0.5 y=1.5
x=-0.5 y=-1.5
x=-1.5 y=-0.5

Answer 3

x^2+2xy+y^2-4=0 (1)
x^2-2xy+y^2-1=0 (2)
(1)-(2)
4xy-3=0
y=3/(4x) (3)
sub(3) into (2)
x^2-2x[3/(4x)]+[3/(4x)]^2-1=0
x^2-5/2+9/(16x^2) =0
16x^4-40x^2+9=0
(4x^2-9)(4x^2-1)=0
x^2=9 or 1
x=3 or -3 or 1 or -1
x=3 ,y=1/4
x=-3, y=-1/4
x=1, y=3/4
x=-1, y=-3/4
(x,y)=(3,1/4) or (-3,-1/4) or (1,3/4) or (-1,-3/4)