y=根号64+(t-10)²+根号16+(t-4)²的最小值为?最小为?
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y=根号64+(t-10)²+根号16+(t-4)²的最小值为?最小为?
y=根号64+(t-10)²+根号16+(t-4)²的最小值为?最小为?
y=根号64+(t-10)²+根号16+(t-4)²的最小值为?最小为?
求导y'=(t-10)/√64+(t-10)²+(t-4)/√16+(t-4)²=0
由于分母都为正数所以4