Преобразование дробных выражений

\( y(x-y) \)

\( \dfrac{x^2 -xy}{y}\cdot \dfrac{y^2}{x}=\dfrac{x^2 -xy}{1}\cdot \dfrac{y}{x}=\dfrac{x(x -y)}{1}\cdot \dfrac{y}{x}= \)

\(= y(x-y) \)

\( \dfrac{a(a+b)}{3b} \)

\( \dfrac{3a}{b^2}\cdot \dfrac{ab+b^2}{9}=\dfrac{a}{b^2}\cdot \dfrac{ab+b^2}{3}=\dfrac{a}{b^2}\cdot \dfrac{b(a+b)}{3}= \)

\(=\dfrac{a(a+b)}{3b} \)

\( \dfrac{2}{m} \)

\( \dfrac{n-m}{mn} \cdot \dfrac{2mn}{mn-m^2}= \dfrac{n-m}{1} \cdot \dfrac{2}{mn-m^2}= \dfrac{n-m}{1} \cdot \dfrac{2}{m(n-m)}=\dfrac{2}{m} \)

\( \dfrac{2(a+b)}{c+d} \)

\( \dfrac{4ab}{cx+dx} \cdot \dfrac{ax+bx}{2ab}= \dfrac{2}{cx+dx} \cdot \dfrac{ax+bx}{1}= \) \(\dfrac{2}{x(c+d)} \cdot \dfrac{x(a+b)}{1}= \dfrac{2(a+b)}{c+d}\)