Question

x^2+2x-y^2+6y-8=0且x+y≠2，求x-y的值

Answer 1

(x+1)^2-(y-3)^2 = 0
|x+1| = |y-3|
1) x+1 = y-3
x-y = -4
2)x+1 = -(y-3)
x+1=-y+3
x+y=2(舍)

Answer 2

x^2+2x-y^2+6y-8 =( x^2 + 2x + 1) - (y^2 - 6y + 9) = (x+1)^2 - (y-3)^2 = (x+1+y-3)(x+1-y+3) = (x+y-2)(x-y+4) = 0
所以 x+y -2 = 0 或者x-y+4 = 0，又因为x+y≠2，所以x-y=-4

Answer 3

x^2 + 2 x - y^2 + 6 y - 8=(4 + x - y) (x + y - 2)=0, x+y≠2 =>4 + x - y = 0
=>x - y = -4