Найдите частное дробей и сократите получившуюся дробь:
Получаем:
\(\displaystyle \frac{2x^{\frac{2}{5}}-2y^{\frac{2}{7}}}{3y^{\frac{2}{7}}}:{\frac{x^{\frac{4}{5}}-y^{\frac{4}{7}}}{y^{\frac{4}{7}}}}=\frac{2x^{\frac{2}{5}}-2y^{\frac{2}{7}}}{3y^{\frac{2}{7}}}\cdot{\frac{y^{\frac{4}{7}}}{x^{\frac{4}{5}}-y^{\frac{4}{7}}}}=\frac{\left(2x^{\frac{2}{5}}-2y^{\frac{2}{7}}\right)y^{\frac{4}{7}}}{3y^{\frac{2}{7}}\left(x^{\frac{4}{5}}-y^{\frac{4}{7}}\right)}{ \small .}\)
Чтобы сократить дробь, разложим выражения \(\displaystyle \color{blue}{2x^{\frac{2}{5}}-2y^{\frac{2}{7}}}\) и \(\displaystyle \color{green}{x^{\frac{4}{5}}-y^{\frac{4}{7}}}\) на множители:
- \(\displaystyle \color{blue}{2x^{\frac{2}{5}}-2y^{\frac{2}{7}}=2\left(x^{\frac{2}{5}}-y^{\frac{2}{7}}\right)}\small,\)
- \(\displaystyle \color{green}{x^{\frac{4}{5}}-y^{\frac{4}{7}}=\left(x^{\frac{2}{5}}-y^{\frac{2}{7}}\right)\left(x^{\frac{2}{5}}+y^{\frac{2}{7}}\right)}\small. \)
Подставляя, получаем:
\(\displaystyle \frac{\color{blue}{\left(2x^{\frac{2}{5}}-2y^{\frac{2}{7}}\right)}y^{\frac{4}{7}}}{3y^{\frac{2}{7}}\color{green}{\left(x^{\frac{4}{5}}-y^{\frac{4}{7}}\right)}}=\frac{\color{blue}{2\left(x^{\frac{2}{5}}-y^{\frac{2}{7}}\right)}y^{\frac{4}{7}}}{3y^{\frac{2}{7}}\color{green}{\left(x^{\frac{2}{5}}-y^{\frac{2}{7}}\right)\left(x^{\frac{2}{5}}+y^{\frac{2}{7}}\right)}}{ \small .}\)
\(\displaystyle \frac{{2\left(x^{\frac{2}{5}}-y^{\frac{2}{7}}\right)}y^{\frac{4}{7}}}{3y^{\frac{2}{7}}{\left(x^{\frac{2}{5}}-y^{\frac{2}{7}}\right)\left(x^{\frac{2}{5}}+y^{\frac{2}{7}}\right)}}= \frac{2y^{\frac{2}{7}}}{3\left(x^{\frac{2}{5}}+y^{\frac{2}{7}}\right)}{\small .}\)
Ответ: \(\displaystyle \frac{2y^{\frac{2}{7}}}{3\left(x^{\frac{2}{5}}+y^{\frac{2}{7}}\right)}{\small .}\)