128. Представьте в виде дроби: a) \(\frac{mx^2-my^2}{2m+8}\) \(\cdot\) \(\frac{3m+12}{my+mx}\); б) \(\frac{ax+ay}{x^2-2xy+y^2}\) \(\cdot\) \(\frac{x^2 - xy}{7x+7y}\); в) \(\frac{x^3 - y^3}{x+y}\) \(\cdot\) \(\frac{x^2-y^2}{x^2 + xy + y^2}\); г) \(\frac{a^2-1}{a^3+1}\) \(\cdot\) \(\frac{a^2-a+1}{a^2+2a +1}\)

a) \(\frac{mx^2-my^2}{2m+8}\) \(\cdot\) \(\frac{3m+12}{my+mx}\) = \(\frac{m(x^2-y^2)}{2(m+4)}\) \(\cdot\) \(\frac{3(m+4)}{m(y+x)}\) = \(\frac{m(x-y)(x+y)}{2(m+4)}\) \(\cdot\) \(\frac{3(m+4)}{m(x+y)}\) = \(\frac{(x-y) \cdot 3}{2}\) = \(\frac{3(x-y)}{2}\)

Ответ: \(\frac{3(x-y)}{2}\)

б) \(\frac{ax+ay}{x^2-2xy+y^2}\) \(\cdot\) \(\frac{x^2 - xy}{7x+7y}\) = \(\frac{a(x+y)}{(x-y)^2}\) \(\cdot\) \(\frac{x(x - y)}{7(x+y)}\) = \(\frac{a \cdot x}{7 \cdot (x-y)}\) = \(\frac{ax}{7(x-y)}\)

Ответ: \(\frac{ax}{7(x-y)}\)

в) \(\frac{x^3 - y^3}{x+y}\) \(\cdot\) \(\frac{x^2-y^2}{x^2 + xy + y^2}\) = \(\frac{(x-y)(x^2+xy+y^2)}{x+y}\) \(\cdot\) \(\frac{(x-y)(x+y)}{x^2 + xy + y^2}\) = (x-y)(x-y) = \((x-y)^2\)

Ответ: \((x-y)^2\)

г) \(\frac{a^2-1}{a^3+1}\) \(\cdot\) \(\frac{a^2-a+1}{a^2+2a +1}\) = \(\frac{(a-1)(a+1)}{(a+1)(a^2-a+1)}\) \(\cdot\) \(\frac{a^2-a+1}{(a+1)^2}\) = \(\frac{a-1}{(a+1)^2}\)

Ответ: \(\frac{a-1}{(a+1)^2}\)