$\frac{a^2 + b^2}{ab} * (\frac{6a + b}{a^2 - b^2} : \frac{6a^3 + b^3 + a^2b + 6ab^2}{2ab^2 - 2a^2b} + \frac{a + b}{a^2 + b^2}) = \frac{a^2 + b^2}{ab(a + b)}$
$\frac{a^2 + b^2}{ab} * (\frac{6a + b}{a^2 - b^2} : \frac{6a^3 + b^3 + a^2b + 6ab^2}{2ab^2 - 2a^2b} + \frac{a + b}{a^2 + b^2}) = \frac{a^2 + b^2}{ab} * (\frac{6a + b}{(a - b)(a + b)} * \frac{2ab(b - a)}{6a(a^2 + b^2) + b(a^2 + b^2)} + \frac{a + b}{a^2 + b^2}) = \frac{a^2 + b^2}{ab} * (\frac{6a + b}{(a - b)(a + b)} * \frac{-2ab(a - b)}{(a^2 + b^2)(6a + b)} + \frac{a + b}{a^2 + b^2}) = \frac{a^2 + b^2}{ab} * (\frac{-2ab}{(a + b)(a^2 + b^2)} + \frac{a + b}{a^2 + b^2}) = \frac{a^2 + b^2}{ab} * \frac{-2ab + (a + b)(a + b)}{(a + b)(a^2 + b^2)} = \frac{-2ab + a^2 + 2ab + b^2}{ab(a + b)} = \frac{a^2 + b^2}{ab(a + b)}$
Тождество доказано.

$(\frac{x}{xy + y^2} - \frac{x^2 + y^2}{x^3 - xy^2} + \frac{y}{x^2 - xy}) : \frac{x^2 - 2xy + y^2}{x^3 + y^3} = \frac{x^2 - xy + y^2}{y(x - y)}$
$(\frac{x}{xy + y^2} - \frac{x^2 + y^2}{x^3 - xy^2} + \frac{y}{x^2 - xy}) : \frac{x^2 - 2xy + y^2}{x^3 + y^3} = (\frac{x}{y(x + y)} - \frac{x^2 + y^2}{x(x^2 - y^2)} + \frac{y}{(x - y)}) : \frac{(x - y)^2}{(x + y)(x^2 - xy + y^2)} = (\frac{x}{y(x + y)} - \frac{x^2 + y^2}{x(x - y)(x + y)} + \frac{y}{x(x - y)}) * \frac{(x + y)(x^2 - xy + y^2)}{(x - y)^2} = \frac{x^2(x - y) - y(x^2 + y^2) + y^2(x + y)}{xy(x - y)(x + y)} * \frac{(x + y)(x^2 - xy + y^2)}{(x - y)^2} = \frac{x^3 - x^2y - x^2y - y^3 + xy^2 + y^3}{xy(x - y)(x + y)} * \frac{(x + y)(x^2 - xy + y^2)}{(x - y)^2} = \frac{x^3 - 2x^2y + xy^2}{xy(x - y)} * \frac{x^2 - xy + y^2}{(x - y)^2} = \frac{x(x^2 - 2xy + y^2)}{xy(x - y)} * \frac{x^2 - xy + y^2}{(x - y)^2} = \frac{x^2 - xy + y^2}{y(x - y)}$
Тождество доказано.

$(\frac{2x^2y + 2xy^2}{7x^3 + x^2y + 7xy^2 + y^3} * \frac{7x + y}{x^2 - y^2} + \frac{x - y}{x^2 + y^2}) * (x^2 - y^2) = x + y$
$(\frac{2x^2y + 2xy^2}{7x^3 + x^2y + 7xy^2 + y^3} * \frac{7x + y}{x^2 - y^2} + \frac{x - y}{x^2 + y^2}) * (x^2 - y^2) = (\frac{2xy(x + y)}{7x(x^2 + y^2) + y(x^2 + y^2)} * \frac{7x + y}{(x - y)(x + y)} + \frac{x - y}{x^2 + y^2}) * (x - y)(x + y) = (\frac{2xy(x + y)}{(x^2 + y^2)(7x + y)}) * \frac{7x + y}{(x - y)(x + y)} + \frac{x - y}{x^2 + y^2}) * (x - y)(x + y) = (\frac{2xy}{(x - y)(x^2 + y^2)} + \frac{x - y}{x^2 + y^2}) * (x - y)(x + y) = \frac{2xy + (x - y)(x - y)}{(x - y)(x^2 + y^2)} * (x - y)(x + y) = \frac{2xy + x^2 - 2xy + y^2}{x^2 + y^2} * \frac{x + y}{1} = 1 * (x + y) = x + y$
Тождество доказано.

$(\frac{5}{a^2 - 2a - ax + 2x} - \frac{1}{8 - 8a + 2a^2} * \frac{20 - 10a}{x - 2}) : \frac{25}{x^3 - 8} = \frac{x^2 + 2x + 4}{5(a - x)}$
$(\frac{5}{a^2 - 2a - ax + 2x} - \frac{1}{8 - 8a + 2a^2} * \frac{20 - 10a}{x - 2}) : \frac{25}{x^3 - 8} = (\frac{5}{a(a - x) - 2(a - x)} - \frac{1}{2(4 - 4a + a^2)} * \frac{10(2 - a)}{x - 2}) : \frac{25}{(x - 2)(x^2 + 2x + 4)} = (\frac{5}{(a - x)(a - 2)} - \frac{1}{2(2 - a)^2} * \frac{10(2 - a)}{x - 2}) * \frac{(x - 2)(x^2 + 2x + 4)}{25} = (\frac{5}{(a - x)(a - 2)} - \frac{5}{(2 - a)(x - 2)}) * \frac{(x - 2)(x^2 + 2x + 4)}{25} = (\frac{5}{(a - x)(a - 2)} + \frac{5}{(x - 2)(a - 2)}) * \frac{(x - 2)(x^2 + 2x + 4)}{25} = \frac{5(x - 2) + 5(a - x)}{(a - x)(a - 2)(x - 2)} * \frac{(x - 2)(x^2 + 2x + 4)}{25} = \frac{5x - 10 + 5a - 5x}{(a - x)(a - 2)} * \frac{x^2 + 2x + 4}{25} = \frac{5(a - 2)(x^2 + 2x + 4)}{(a - x)(a - 2)25} = \frac{x^2 + 2x + 4}{5(a - x)}$
Тождество доказано.

$(\frac{3a}{9 - 3x - 3a + ax} - \frac{1}{a^2 - 9} : \frac{x - a}{3a^2 + 9a}) * \frac{x^3 - 27}{3a} = \frac{x^2 + 3x + 9}{a - x}$
$(\frac{3a}{9 - 3x - 3a + ax} - \frac{1}{a^2 - 9} : \frac{x - a}{3a^2 + 9a}) * \frac{x^3 - 27}{3a} = (\frac{3a}{3(3 - x) - a(3 - x)} - \frac{1}{(a - 3)(a + 3)} * \frac{3a(a + 3)}{x - a}) * \frac{(x - 3)(x^2 + 3x + 9)}{3a} = (\frac{3a}{(3 - x)(3 - a)} - \frac{3a}{(a - 3)(x - a)}) * \frac{(x - 3)(x^2 + 3x + 9)}{3a} = (\frac{3a}{(3 - x)(3 - a)} + \frac{3a}{(3 - a)(x - a)}) * \frac{(x - 3)(x^2 + 3x + 9)}{3a} = \frac{3a(x - a) + 3a(3 - x)}{(3 - x)(3 - a)(x - a)} * \frac{(x - 3)(x^2 + 3x + 9)}{3a} = \frac{3a(x - a + 3 - x)}{-(x - 3)(3 - a)(x - a)} * \frac{(x - 3)(x^2 + 3x + 9)}{3a} = \frac{(3 - a)(x^2 + 3x + 9)}{-(3 - a)(x - a)} = \frac{x^2 + 3x + 9}{a - x}$
Тождество доказано.