129. Упростите выражение: a) \(\frac{x^2-10x+25}{3x+12}\) \(\cdot\) \(\frac{x^2-16}{2x-10}\); б) \(\frac{1-a^2}{4a+8b}\) \(\cdot\) \(\frac{a^2+4ab + 4b^2}{3-3a}\); в) \(\frac{y^2 - 25}{y^2 +12y +36}\) \(\cdot\) \(\frac{3y+18}{2y+10}\); г) \(\frac{b^3 +8}{18b^2+27b}\) \(\cdot\) \(\frac{b^2 - 2b+4}{b^2 - 2b+4}\)

a) \(\frac{x^2-10x+25}{3x+12}\) \(\cdot\) \(\frac{x^2-16}{2x-10}\) = \(\frac{(x-5)^2}{3(x+4)}\) \(\cdot\) \(\frac{(x-4)(x+4)}{2(x-5)}\) = \(\frac{(x-5)(x-4)}{3 \cdot 2}\) = \(\frac{(x-5)(x-4)}{6}\)

Ответ: \(\frac{(x-5)(x-4)}{6}\)

б) \(\frac{1-a^2}{4a+8b}\) \(\cdot\) \(\frac{a^2+4ab + 4b^2}{3-3a}\) = \(\frac{(1-a)(1+a)}{4(a+2b)}\) \(\cdot\) \(\frac{(a+2b)^2}{3(1-a)}\) = \(\frac{(1+a)(a+2b)}{4 \cdot 3}\) = \(\frac{(1+a)(a+2b)}{12}\)

Ответ: \(\frac{(1+a)(a+2b)}{12}\)

в) \(\frac{y^2 - 25}{y^2 +12y +36}\) \(\cdot\) \(\frac{3y+18}{2y+10}\) = \(\frac{(y-5)(y+5)}{(y+6)^2}\) \(\cdot\) \(\frac{3(y+6)}{2(y+5)}\) = \(\frac{(y-5) \cdot 3}{2(y+6)}\) = \(\frac{3(y-5)}{2(y+6)}\)

Ответ: \(\frac{3(y-5)}{2(y+6)}\)

г) \(\frac{b^3 +8}{18b^2+27b}\) \(\cdot\) \(\frac{b^2 - 2b+4}{b^2 - 2b+4}\) = \(\frac{(b+2)(b^2 - 2b+4)}{9b(2b+3)}\) \(\cdot\) 1 = \(\frac{(b+2)(b^2 - 2b+4)}{9b(2b+3)}\)

Ответ: \(\frac{(b+2)(b^2 - 2b+4)}{9b(2b+3)}\)