Решение:
- \(\frac{3a-3b}{6a} = \frac{3(a-b)}{6a} = \frac{a-b}{2a}\)
- \(\frac{7x}{7x+14y} = \frac{7x}{7(x+2y)} = \frac{x}{x+2y}\)
- \(\frac{2a^2b}{2a^3b-6a^2b} = \frac{2a^2b}{2a^2b(a-3)} = \frac{1}{a-3}\)
- \(\frac{3m-3n}{6(m+n)} = \frac{3(m-n)}{6(m+n)} = \frac{m-n}{2(m+n)}\)
- \(\frac{3x+6y}{5x+10y} = \frac{3(x+2y)}{5(x+2y)} = \frac{3}{5}\)
- \(\frac{5n-m}{5n^2-mn} = \frac{5n-m}{n(5n-m)} = \frac{1}{n}\)
- \(\frac{(2c+3d)^3}{2ac+3ad} = \frac{(2c+3d)^3}{a(2c+3d)} = \frac{(2c+3d)^2}{a}\)
- \(\frac{3x^2y-x^2}{3y-z} \text{ (невозможно сократить, предполагая, что в знаменателе } (3y-z)^1)\)
- \(\frac{xy+xz}{xy} = \frac{x(y+z)}{xy} = \frac{y+z}{y}\)
- \(\frac{9a^2(a-2b)}{3a-6b} = \frac{9a^2(a-2b)}{3(a-2b)} = 3a^2\)
- \(\frac{7a-14b}{(a-2b)^2} = \frac{7(a-2b)}{(a-2b)^2} = \frac{7}{a-2b}\)
- \(\frac{5a(c+6d)^2}{5ac+30ad} = \frac{5a(c+6d)^2}{5a(c+6d)} = c+6d\)
Ответ: \(\frac{a-b}{2a}\), \(\frac{x}{x+2y}\), \(\frac{1}{a-3}\), \(\frac{m-n}{2(m+n)}\), \(\frac{3}{5}\), \(\frac{1}{n}\), \(\frac{(2c+3d)^2}{a}\), \(\frac{3x^2y-x^2}{3y-z}\), \(\frac{y+z}{y}\), \(3a^2\), \(\frac{7}{a-2b}\), \(c+6d\).