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∵x、y均為正實數,且12+x+12+y=13,進一步化簡得xy-x-y-8=0.x+y=xy-8≥2xy,令t=xy,t2-2t-8≥0,∴t≤-2(舍去),或t≥4,即xy≥4,化簡可得 ;xy≥16,∴xy的最小值為16.
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