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+y│+│y+1│+│2y-x-4│的取值范围是
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.
答:设y/x=t,代入原方程得x^2+(tx)^2-4x+1=0 ==> (1+t^2)x^2-4x+1=0,其判别式不小于0,故(-4)^2-4(1+t^2)>=...详情>>
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