2y2-2 (1-x) y+2x2-2x+0.5 = 0, ∫y∈R, so δ≥ 0.
∴ 4 (1-x) 2-4× 2 (2x2-2x+0.5) ≥ 0, and 0≤x≤2\3, ∴ x ∈ [0 0,2 \ 3]
Y, z ∈ [0 0,2 \ 3] can be obtained in the same way.