设x，y> 0，且xy-x-y=1,则 A。x+y ≥2(2√2+1) B.xy≤√2+1 C.x+y≤(√2+1)^2 D.xy≥2(√2+1)
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因为x+y=(xy)^0.5所以xy-1=2(xy)^0.5设(xy)^0.5=tt^2-1=2t (t0)解得t=1+2^0.5即xy=3+2*2^0.5 BD排除因为(x+y)/2=(xy)^0.5所以(x+y+1)^0.50)(p+1)^0.5=2(1+2^0.5) 选A
设x，y> 0，且xy-x-y=1,则 A。x+y ≥2(2√2+1) B.xy≤√2+1 C.x+y≤(√2+1)^2 D.xy≥2(√2+1)
因为x+y=(xy)^0.5所以xy-1=2(xy)^0.5设(xy)^0.5=tt^2-1=2t (t0)解得t=1+2^0.5即xy=3+2*2^0.5 BD排除因为(x+y)/2=(xy)^0.5所以(x+y+1)^0.50)(p+1)^0.5=2(1+2^0.5) 选A