【答案】
分析:由于x,y,z滿足方程x
2+(y-2)
2+(z+2)
2=2,在空間直角坐標中,它表示球心在A(0,2,-2)半徑為r=
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的球,球面上一點P(x,y,z)到原點的距離為:
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,利用幾何圖形的特點即可求得
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的最大值是OA+r.
解答:解:因x,y,z滿足方程x
2+(y-2)
2+(z+2)
2=2,
在空間直角坐標中,它表示球心在A(0,2,-2)半徑為r=

的球,
球面上一點P(x,y,z)到原點的距離為:
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則
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的最大值是即為:
OA+r=
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+
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=3
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.
故選A.
點評:本題主要考查隨時隨最值的求法,解答關鍵是數(shù)形結(jié)合,把滿足方程x
2+(y-2)
2+(z+2)
2=2點P(x,y,z)看成是球心在A(0,2,-2)半徑為r=

的球.