y=(x+1)/(x²+5x+6) 设x+1=t>0,那么x=t-1∴y=t/[(t-1)²+5(t-1)+6] =t/(t²+3t+2)取倒数1/y=t+2/t+3∵t>0∴t+2/t≥2√(t*2/t)=2√2当且仅当t=2/t,t=√2时取等号∴1/y≥3+2√2∴y≤1/(3+2√2)=3-2...