阅读例题,解答下题.
范例:解方程:x2 + ∣x +1∣﹣1= 0
解:(1)当 x+1 ≥ 0,即x ≥﹣1时,
x2 + x +1﹣1= 0
x2 + x = 0
解得 x 1 = 0,x2 =﹣1
(2)当 x+1 < 0,即x <﹣1时,
x2 ﹣( x +1)﹣1= 0
x2﹣x ﹣2= 0
解得x 1 =﹣1,x2 = 2
∵ x <﹣1,∴ x 1 =﹣1,x2 = 2 都舍去.
综上所述,原方程的解是x1 = 0,x2 =﹣1
依照上例解法,解方程:x2﹣2∣x-2∣-4 = 0