Найдите произведение дробей и сократите получившуюся дробь:
Получаем:
\(\displaystyle \frac{2x^{\frac{1}{7}}-2y^{\frac{1}{5}}}{y^{\frac{1}{5}}}\cdot\frac{3y^{\frac{2}{5}}}{x^{\frac{2}{7}}-y^{\frac{2}{5}}}=\frac{3y^{\frac{2}{5}}\left(2x^{\frac{1}{7}}-2y^{\frac{1}{5}}\right)}{y^{\frac{1}{5}}\left(x^{\frac{2}{7}}-y^{\frac{2}{5}}\right)}{ \small .}\)
Разложим \(\displaystyle 2x^{\frac{1}{7}}-2y^{\frac{1}{5}}\) и \(\displaystyle x^{\frac{2}{7}}-y^{\frac{2}{5}}\) на множители:
\(\displaystyle 2x^{\frac{1}{7}}-2y^{\frac{1}{5}}=2\left(x^{\frac{1}{7}}-y^{\frac{1}{5}}\right)\)
и
\(\displaystyle x^{\frac{2}{7}}-y^{\frac{2}{5}}=\left(x^{\frac{1}{7}}\right)^2-\left(y^{\frac{1}{5}}\right)^2=\left(x^{\frac{1}{7}}-y^{\frac{1}{5}}\right)\left(x^{\frac{1}{7}}+y^{\frac{1}{5}}\right)\small. \)
Подставляя, получаем:
\(\displaystyle \frac{3y^{\frac{2}{5}}{\left(2x^{\frac{1}{7}}-2y^{\frac{1}{5}}\right)}}{y^{\frac{1}{5}}{\left(x^{\frac{2}{7}}-y^{\frac{2}{5}}\right)}}=\frac{3y^{\frac{2}{5}} \cdot {2\left(x^{\frac{1}{7}}-y^{\frac{1}{5}}\right)}}{y^{\frac{1}{5}}{\left(x^{\frac{1}{7}}-y^{\frac{1}{5}}\right)\left(x^{\frac{1}{7}}+y^{\frac{1}{5}}\right)}}=\frac{6y^{\frac{2}{5}} \left(x^{\frac{1}{7}}-y^{\frac{1}{5}}\right)}{y^{\frac{1}{5}}{\left(x^{\frac{1}{7}}-y^{\frac{1}{5}}\right)\left(x^{\frac{1}{7}}+y^{\frac{1}{5}}\right)}}{ \small .}\)
Сокращая, получаем:
\(\displaystyle \begin{aligned}\frac{6{\color{green}{y^{\frac{2}{5}}}} \color{blue}{\left(x^{\frac{1}{7}}-y^{\frac{1}{5}}\right)}}{{\color{green}{y^{\frac{1}{5}}}}{\color{blue}{\left(x^{\frac{1}{7}}-y^{\frac{1}{5}}\right)}\left(x^{\frac{1}{7}}+y^{\frac{1}{5}}\right)}}=\frac{6\color{green}{y^{\frac{2}{5}-\frac{1}{5}}}{\cancel{\color{blue}{\left(x^{\frac{1}{7}}-y^{\frac{1}{5}}\right)}}}}{\cancel{\color{blue}{\left(x^{\frac{1}{7}}-y^{\frac{1}{5}}\right)}}\left(x^{\frac{1}{7}}+y^{\frac{1}{5}}\right)}=\frac{6y^{\frac{1}{5}}}{x^{\frac{1}{7}}+y^{\frac{1}{5}}} {\small .}\end{aligned}\)
Ответ: \(\displaystyle \frac{6y^{\frac{1}{5}}}{x^{\frac{1}{7}}+y^{\frac{1}{5}}}{\small .}\)