原式=[(a+b)²/(a-b)²][2(a-b)/3(a+b)]-[a²/(a+b)(a-b)](b/a)
=[2(a+b)/3(a-b)]-[ab/(a+b)(a-b)]
=[(2a+2b)(a+b)-3ab]/3(a+b)(a-b)
=(2a²+ab+2b²)/(3a²-3b²)
希望能解决您的问题.