Question

Помогите пожалуйста, заранее спасибо!!!

Answer 1

$\frac{y^2}{xy-5x+y-5}+\frac{2}{x+1}=\frac{y^2}{x(y-5)+1(y-5)}+\frac{2}{x+1} =\frac{y^2}{(x+1)(y-5)}+\frac{2}{x+1}=\\\\=\frac{y^2+2(y-5)}{(x+1)(y-5)} =\frac{y^2+2y-10}{(x+1)(y-5)}$


$\frac{a^2+3b}{a^2-ab+3a-3b}-\frac{ab-3a}{a^2+ab+3a+3b}+1 =\frac{a^2+3b}{a(a-b)+3(a-b)}-\frac{a(b-3)}{a(a+b)+3(a+b)}+1=\\\\ =\frac{a^2+3b}{(a+3)(a-b)}-\frac{a(b-3)}{(a+3)(a+b)}+1 =\frac{(a^2+3b)(a+b)-a(b-3)(a-b)+(a+3)(a-b)(a+b)}{(a+3)(a-b)(a+b)}\\\\ =\frac{a^3+a^2b+3ab+3b^2-a^2b+3a^2+ab^2-3ab+a^3+3a^2-ab^2-3b^2}{(a+3)(a-b)(a+b)}=\\\\ =\frac{2a^3+6a^2}{(a+3)(a-b)(a+b)}=\frac{2a^2(a+3)}{(a+3)(a-b)(a+b)} =\frac{2a^2}{(a-b)(a+b)}=\frac{2a^2}{a^2-b^2}$