Question

Задание 2.ПЖ!!!Помогите

Answer 1

а)  $\frac{3an}{2(a+n)^2} * \frac{a^2-n^2}{3n} = \frac{3an(a^2-n^2)}{2(a+n)^2*3n} = \frac{a(a^2-n^2)}{2(a+n)^2} = \frac{a(a+n)(a-n)}{2(a+n)^2} = \frac{a(a-n)}{2(a+n)}.$

б) $(a+x)*\frac{ax}{a^3+x^3} = \frac{ax(a+x)}{(a+x)(a^2-ax+x^2)} = \frac{ax}{a^2-ax+x^2}.$

в) $\frac{5}{x+y}:\frac{3ab}{a-b} = \frac{5(a-b)}{3ab(x+y)} = \frac{5a-5b}{3ab(x+y)} = \frac{5a}{3ab(x+y)} - \frac{5b}{3ab(x+y)} = \frac{5}{3b(x+y)} - \frac{5}{3a(x+y)}.$

г) $\frac{2xy}{5(x-y)}:\frac{3x^2}{(y-x)^2} = \frac{2xy(y-x)^2}{5(x-y)*3x^2} = \frac{2xy(x-y)^2}{15x^2(x-y)} = \frac{2y(x-y)}{15x} = \frac{2xy-2y^2}{15x} =\\= \frac{2xy}{15x} - \frac{2y^2}{15x} = \frac{2y}{15} - \frac{2y^2}{15x}.$

Answer 2

решение смотри внизу