若正数a,b满足aˆ2/(aˆ4+aˆ2+1)=1/24,bˆ3/(bˆ6+bˆ3+1)=1/19,则ab/(a^2+a+1)(b^2+b+1)=(),+

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若正数a,b满足aˆ2/(aˆ4+aˆ2+1)=1/24,bˆ3/(bˆ6+bˆ3+1)=1/19,则ab/(a^2+a+1)(b^2+b+1)=(),+
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若正数a,b满足aˆ2/(aˆ4+aˆ2+1)=1/24,bˆ3/(bˆ6+bˆ3+1)=1/19,则ab/(a^2+a+1)(b^2+b+1)=(),+
若正数a,b满足aˆ2/(aˆ4+aˆ2+1)=1/24,bˆ3/(bˆ6+bˆ3+1)=1/19,
则ab/(a^2+a+1)(b^2+b+1)=(),
+

若正数a,b满足aˆ2/(aˆ4+aˆ2+1)=1/24,bˆ3/(bˆ6+bˆ3+1)=1/19,则ab/(a^2+a+1)(b^2+b+1)=(),+
因为aˆ2/(aˆ4+aˆ2+1)=1/24,(aˆ4+aˆ2+1)/a^2=24,a^2+2+(1/a)^2=25,(a+1/a)^2=25
a+1/a=5
bˆ3/(bˆ6+bˆ3+1)=1/19,b^3+1+(1/b)^3=19,(b+1/b)(b^2-1+1/b^2)=18,(b+1/b)((b+1/b)^2-3)=18
b+1/b=3,
所以原式=1/(a+1/a+1)(b+1/b+1)=1/24