初二数学问题【分式】，急啊，，追加【20分】！

a/b=3/2=1.5
a=1.5b

a/(a+b)+b/(a-b)-b²/(a²-b²)
=a(a-b)/(a+b)(a-b)+b(a+b)/(a+b)(a-b)-b²/(a+b)(a-b)
=(a²-ab+ab+b²-b²)/(a+b)(a-b)
=a²/(a+b)(a-b)
=2.25b²/(1.5b+b)(1.5b-b)
=2.25b²/1.25b²
=9/5

a/a+b + b/a-b - b²/a²-b²

=(a/b)/(a/b+1)+1/(a/b-1)-1/[(a/b)^2-1]
=(3/2)/(3/2+1)+1/(3/2-1)-1/[(3/2)^2-1]
=3/5+2-4/5
=9/5

a/b=3/2

设a=3k,b=2k

a/a+b + b/a-b - b²/a²-b²

=3/(3+2)+2/(3-2)-4/(9-4)

=9/5

a/b=3/2 b/a=2/3
所以a/(a+b)=1/[(a+b)/a]=1/(1+b/a)=3/5
b/(a-b)=1/[(a-b)/b]=1/(a/b-1)=2
b²/(a²-b²)=1/[(a²-b²)/b²]=1/[(a/b)²-1]
=4/5
a/(a+b) + b/(a-b) - b²/(a²-b²)
=3/5+2-4/5
=9/5

因为a/b=3/2 所以a=3 b=2
然后再化简:a/a+b + b/a-b - b²/a²-b²
=a(a-b)+b