令(x-y)/(x+2y+3)> 0 ,有(x-y)/(x+2y+3)+(x+2)/(x+2y+3)+(5y+4)/(x+2y+3)=2 ,f(x,y)的最小值为2/3(x=-8,y=-2).若(x-y)/(x+2y+3)〈 0,设(x-y)〈0,则(x+2y+3)> 0,y〉x,令y=x+t,有t>0,(y-x)/(x+2y+3)+(x+2)/(x+2y+3)+(5y+4)/(x+2y+3)=2+2t/(3x+2t+3)>2,f(x,y)的最小值大于2/3.设y0,(y-x)/(x+2y+3)+(x+2)/(x+2y+3)+(5y+4)/(x+2y+3)=2-2s/(3y+s+3)>2,f(x,y)的最小值大于2/3.所以f(x,y)的最小值为2/3(x=-8,y=-2).
求f(x,y)=,max{|(x-y)/(x+2y+3)|,(x+2)/(x+2y+3),(5y+4)/(x+2y+3)
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