Problem:
If x=baβ,aξ =b and bξ =0, then aβba+bβ=
Answer Choices:
A. x+1xβ
B. xβ1x+1β
C. 1
D. xβx1β
E. x+x1β
Solution:
Since a=bx,aβba+bβ=bxβbbx+bβ=b(xβ1)b(x+1)β=xβ1x+1β. To show that the other four choices are incorrect, let a=2 and b=1.
OR
Dividing numerator and denominator by b shows that
aβba+bβ=baββ1baβ+1β=xβ1x+1β
Note. More generally, if tsβ=vuβξ =1 then sβts+tβ=uβvu+vβ, since both sides of the last equation equal vuβtβtvuβt+tβ.