已知1/a+1/b=2/3,求代数式[a/(a+b)-b/(b-a)-2ab/(a2-b2)除以(1/a-1/b)的值

1/a+1/b=2/3[a/(a+b)-b/(b-a)-2ab/(a2-b2)] /(1/a-1/b)= [a/(a+b)+b/(a-b)-2ab/(a2-b2)] / [(b-a)/(ab)]= { [a(a-b)+b(a+b)-2ab]/(a2-b2)] } / [-(a-b)/(ab)]= { a^2-ab+ab+b^2-2ab]/(a2-b2)] } / [-(a-b)/(ab)]= -(a-b)^2/(a+b)(a-b) * ab/(a-b)= -ab/(a+b) = -1/(1/a+1/b)= -1/(2/3)= -3/2