Question

已知：a平方-3a+1=0求2倍的a平方-5a-2+(a2+1)分之3得多少?

Answer 1

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

故原式=3-4=-1