f.5數學一問

case 1:
a=1/1-b/b-1, =1/(1-b/b-1)
b-1/1-b=a, therefore b-1=a-ab,
b+ab=a+1
b(1+a)=a+1, therefore b=1

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

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30!!!!!!