若正数a.b满足a^2/(a^4+a^2+1)=1/24,b^3/(b^6+b^3+1)=1/19,则ab/(a^2+a+1)(b^2+b+1)=

题目

若正数a.b满足a^2/(a^4+a^2+1)=1/24,b^3/(b^6+b^3+1)=1/19,则ab/(a^2+a+1)(b^2+b+1)=

答案

因为,a^2/(a^4+a^2+1)=1/24
所以,24=(a^4+a^2+1)/a^2=a^2+1+1/a^2=(a+1/a)^2-1
由于a>0 所以,a+1/a=√(24+1)=5
又因为b^3/(b^6+b^3+1)=1/19
所以,19=b^3+1+1/b^3
令t=b+1/b可知,t≥2
且有,t^3=b^3+3b+3/b+1/b^3=b^3+1+1/b^3-1+3(b+1/b)
即t^3=19-1+3t
t^3-3t-18=0
t^3-3t^2+3t^3-9t+6t-18=0
(t-3)(t^2+3t+6)=0
由于t^2+3t+6>0,所以t=3
ab/(a^2+a+1)(b^2+b+1)取倒数为
(a^2+a+1)(b^2+b+1)/ab=(a+1+1/a)(b+1+1/b)
=(5+1)(3+1)=24
所以,ab/(a^2+a+1)(b^2+b+1)=1/24