P-Q=a^2b^2+5-2ab+a^2+4a
=(a^2b^2-2ab+1)+(a^2+4a+4)=(ab-1)^2+(a+2)^2
當(dāng)且僅當(dāng)ab=1且a=-2,即a=-2,b=-1/2時取到等號
∴a=-2,b=-1/2
設(shè)P=a^2b^2+5,Q=2ab-a^2-4a,若P=Q,則實(shí)數(shù)a= ,b=
設(shè)P=a^2b^2+5,Q=2ab-a^2-4a,若P=Q,則實(shí)數(shù)a= ,b=
數(shù)學(xué)人氣：409 ℃時間：2019-11-10 09:49:17
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