2倍的a平方-5a-2+(a2+1)分之3 =2(a^2-3a+1)+a-4+3/(a^2+1) ∵a平方-3a+1=0,∴原式=2×0+a-4+3/(a^2+1)=a-4+3/(a^2+1)∵a^2-3a+1=0∴a^2+1=3a代入原式=a-4+3/3a=a-4+1/a=(a+1/a)-4将a^2-3a+1=0两边除以a得a+1/a=3故原...