2 вариант (ответы с решением) 1 и 3 задание 2 не нужно.

Решите задачу:

$\frac{24ab^{2} }{18a^{4}b^{2} } = \frac{6*4ab^2}{6*3*a^4b^2} =\frac{4}{3a^{3} } \\ \\ \frac{10 x^{2}-15 xy}{12 y^{2}-8xy } = \frac{5x(2x-3y)}{4y(3y-2x)} = \frac{5x}{-4y} =- \frac{5x}{4y}$

$\frac{8x^2y^2}{x^2-y^2} : ( \frac{x}{x-y} - \frac{x}{x+y} )= \\ = \frac{8x^2y^2}{x^2-y^2} : \frac{x(x+y) -x(x-y) }{(x-y)(x+y)} = \\ =\frac{8x^2y^2}{x^2-y^2} : \frac{x^2+xy -x^2+xy }{x^2-y^2}= \\ = \frac{8x^2y^2 * (x^2-y^2)}{(x^2-y^2) * 2xy}= 4xy$

$\frac{1}{2xy^2} - ( \frac{x}{x-y} - \frac{x}{y-x} ) * ( \frac{1}{ x^{2} } - \frac{2}{xy} + \frac{1}{y^2} )= \\ = \frac{1}{2xy^2} - ( \frac{x}{x-y} + \frac{x}{x-y} ) *( \frac{y^2-2xy+x^2}{ x^{2} y^2} ) = \\ = \frac{1}{2xy^2} - ( \frac{2x}{x-y} ) * ( \frac{-(x-y)^2}{x^2y^2} ) = \\ = \frac{1}{2xy^2} - (- \frac{2x(x-y)}{x^2y^2} ) = \frac{x+ 2*2x(x-y)}{2x^2y^2} = \\ = \frac{x+4x^2-4xy}{2x^2y^2} = \frac{x(1+4x-4y)}{2x^2y^2} = \frac{1+4x-4y}{2xy^2}= \frac{0.5+2x-2y}{xy^2}$