Question

先化简,再求值：1/（a+1）-（a+3）/（a^2-1）*（a^2-2a+1）/（a^2+4a+3）,其中a满足a^2

Answer 1

1/(a+1)-[(a+3)/(a^2-1)]*[(a^2-2a+1)/(a^2+4a+3)]

=1/(a+1)-[(a+3)(a-1)^2]/[(a+1)^2(a-1)(a+3)]

=1/(a+1)-(a-1)/(a+1)^2

=(a+1)/(a+1)^2-(a-1)/(a+1)^2

=[(a+1)-(a-1)]/(a+1)^2

=2/(a+1)^2

又a^2+2a-1=0

a^2+2a+1=2

(a+1)^2=2

所以原式=2/(a+1)^2=2/2=1