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数学
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已知复数z=x+yi,实数x,y满足x≥1,y≤2,x-y≤1,则|z-4|的最小值是
人气:204 ℃ 时间:2020-04-26 11:16:00
解答
这个用作图,x代表横坐标,y是纵坐标
x≥1,y≤2,x-y≤1,可以画出可行域
|z-4|即|(x-4)+yi,|
即求原点到(x-4,y)的距离的最小值作图可知是点(-1,0)可得最小值=1
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