求y=6根号(x2+1)/(x2+4)的最大值

y=6√(x^2+1)/(x^2+4)令t=√(x^2+1)≥1,则y=6√(x^2+1)/(x^2+4)=6t/(t^2+3)利用对勾函数的特性求1/y=t/6+1/(2t)≥2√[t/6*1/(2t)]=1/√3,∴y≤√3取等号的条件：t/6=1/(2t)得到t^2=3(满足t≥1)即x^2+1=3,∴x=±√2∴...