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Решите задачу:

$1)\\\\(\frac{a^2+b^2}{ab} +2)*\frac{ab}{a+b}=(\frac{a^2+b^2+2ab}{ab})*\frac{ab}{a+b}=\\\\=\frac{(a+b)^2}{ab}*\frac{ab}{a+b}=\frac{(a+b)^2*ab}{ab*(a+b)}=\\\\=\frac{a+b}{1}=a+b$

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

$3)\\\\(\frac{x+2}{x^2+2x+4}-\frac{6x-13}{x^3-8})*\frac{2x^2+4x+8}{3-x}=\\\\=(\frac{x+2}{x^2+2x+4}-\frac{6x-13}{(x-2)(x^2+2x+4)})*\frac{2(x^2+2x+4)}{3-x}=\\\\=(\frac{(x+2)*(x-2)-(6x-13)}{(x-2)(x^2+2x+4)})*\frac{2(x^2+2x+4)}{3-x}=\\\\=(\frac{(x^2-4)-6x+13}{(x-2)(x^2+2x+4)})*\frac{2(x^2+2x+4)}{3-x}=\\\\=\frac{x^2-6x+9}{(x-2)(x^2+2x+4)}*\frac{2(x^2+2x+4)}{3-x}=\\\\=\frac{(3-x)^2}{(x-2)(x^2+2x+4)}*\frac{2(x^2+2x+4)}{3-x}=\\\\=\frac{(3-x)^2*2*(x^2+2x+4)}{(x-2)(x^2+2x+4)*(3-x)}=\frac{2*(3-x)}{x-2}$

$4)\\\\(\frac{x-2}{x^2-2x+4}-\frac{2x-5}{x^3+8})*\frac{3x^2-6x+12}{1-x}=\\\\=(\frac{x-2}{x^2-2x+4}-\frac{2x-5}{(x+2)(x^2-2x+)})*\frac{3(x^2-2x+4)}{1-x}=\\\\=(\frac{(x-2)(x+2)-(2x-5)}{(x+2)(x^2-2x+4)})*\frac{3(x^2-2x+4)}{1-x}=\\\\=\frac{(x^2-4)-2x+5}{(x+2)(x^2-2x+4)}*\frac{3(x^2-2x+4)}{1-x}=\\\\=\frac{x^2-2x+1}{(x+2)(x^2-2x+4)}*\frac{3(x^2-2x+4)}{1-x}=\\\\\frac{(1-x)^2*3*(x^2-2x+4)}{(x+2)(x^2-2x+4)*(1-x)}=\frac{3*(1-x)}{x+2}$