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教育/科学

已知实数xy满足x2+y2≤1已知实数x,y满足x2+y2≤1,

2009-09-29 20:06:22s***
已知实数x,y满足x2+y2≤1,则│x+y│+│y+1│+│2y-x-4│的取值范围是已知实数x,y满足x2+y2≤1，则│x+y│+│y+1│+│2y-x-4│的取值范围是已知实数xy满足x2+y2≤1已知实数x,y满足x2+y2≤1,则│x+y│+│y+1│+│2y-x-4│的取值范围是已知实数x,y满足x2+y2≤1，则│x+?

最佳回答

  • 注意到y+1>=0,2y-x-4=0时,-1/√2<=x<=1,原式=x+y+x-y+5=2x+5,故 5-√2<=│x+y│+│y+1│+│2y-x-4│<=7. 当x+y<0时,-1<=y<=1/√2,原式=-2y+5,故 5-√2<=│x+y│+│y+1│+│2y-x-4│<=7.综上 5-√2<=│x+y│+│y+1│+│2y-x-4│<=7.
    2009-09-30 23:52:45
    • 很赞哦！ (227)

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