Решите пожалуйста!!Заранее спасибо))Под г.

[latex]\displaystyle \frac{a^{2}+3b }{a^{2}-ab+3a-3b}- \frac{ab-3a}{a^{2}+ab+3a+3b }+1 = \\ \\ = \frac{a^{2}+3b}{a(a-b)+3(a-b)}- \frac{a(b-3)}{a(a+b)+3(a+b)}+1= \\ \\ = \frac{(a^{2}+3b)(a+b)-a(b-3)(a-b)}{(a+3)(a-b)(a+b)}+1= \\ \\ = \frac{a^{3}+a^{2}b+3ab+3b^{2}-a^{2} b+ab^{2}+3a^{2} -3ab }{(a+3)(a-b)(a+b)}+1= \\ \\ = \frac{a^{3}+ab^{2}+3a^{2}+3b^{2}}{(a+3)(a-b)(a+b)}+1= \frac{a^{2}(a+3)+b^{2}(a+3)}{(a+3)(a-b)(a+b)}+1= \\ \\ = \frac{a^{2}+b^{2}+a^{2}-b^{2}}{a^{2}-b^{2}}= \frac{2a^{2}}{a^{2}-b^{2}}; [/latex]