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yx=y−0x−0,即連接圓上一點與座標原點的直線的斜率,囙此yx的最值即為過原點的直線與圓相切時該直線的斜率.設yx=k,則kx-y=0.由|2k|1+k2=3,得k=±3,故(yx)max=3,(yx)min=-3.故答案為:3
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