y
= (2x² -1)/(x² + 3)
= (2x² + 6 - 7)/(x² + 3)
= (2x² + 6)/(x² + 3) - 7/(x² + 3)
= 2 - 7/(x² + 3)
x² ≥ 0
x² + 3 ≥ 3
-7/3 ≤ -7/(x² + 3) < 0
-1/3 ≤ 2 - -7/(x² + 3) < 2
所以 -1/3 ≤ y < 2
所以值域是 [-1/3 ,2)