设x+2y=t,代入条件式得
(t-2y)y=(t-2y)+y+1
→2y²-(t+1)y+t+1=0
△=(t+1)²-8(t+1)≥0
→(t+1)(t-7)≥0
→t≤-1,t≥7.
故所求最小值为:(x+2y)|min=7.
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