Вопрос:

3. Упростите: а) \(\frac{x^2-y^2}{3xy}\cdot\frac{3y}{x-y}\); б) \(\frac{c^2-49}{10cd}:\frac{2c+14}{5d}\); в) \(\frac{x^2-10x+25}{3x+12}:\frac{2x-10}{x^2-16}\); г) \(\frac{t^3+8}{12t^2+27t}\cdot\frac{4t+9}{t^2-2t+4}\).

Ответ:

а) Разложим разность квадратов: \(x^2-y^2=(x-y)(x+y)\).

\[\frac{(x-y)(x+y)}{3xy}\cdot\frac{3y}{x-y}=\frac{x+y}{x}.\]

б) \(c^2-49=(c-7)(c+7)\), \(2c+14=2(c+7)\).

\[\frac{(c-7)(c+7)}{10cd}\cdot\frac{5d}{2(c+7)}=\frac{c-7}{4c}.\]

в) \(x^2-10x+25=(x-5)^2\), \(3x+12=3(x+4)\), \(x^2-16=(x-4)(x+4)\).

\[\frac{(x-5)^2}{3(x+4)}\cdot\frac{(x-4)(x+4)}{2(x-5)}=\frac{(x-5)(x-4)}{6}.\]

г) \(t^3+8=(t+2)(t^2-2t+4)\), \(12t^2+27t=3t(4t+9)\).

\[\frac{(t+2)(t^2-2t+4)}{3t(4t+9)}\cdot\frac{4t+9}{t^2-2t+4}=\frac{t+2}{3t}.\]

Ответ: а) \(\frac{x+y}{x}\); б) \(\frac{c-7}{4c}\); в) \(\frac{(x-5)(x-4)}{6}\); г) \(\frac{t+2}{3t}\).