Найдите произведение дробей и сократите получившуюся дробь:
Получаем:
\(\displaystyle \frac{a^{\frac{2}{3}}-2a^{\frac{1}{3}}b^{\frac{1}{5}}+b^{\frac{2}{5}}}{2b^{\frac{1}{5}}+3}\cdot \frac{a^{\frac{2}{3}}}{a-a^{\frac{2}{3}}b^{\frac{1}{5}}}=\frac{(a^{\frac{2}{3}}-2a^{\frac{1}{3}}b^{\frac{1}{5}}+b^{\frac{2}{5}})\cdot a^{\frac{2}{3}}}{(2b^{\frac{1}{5}}+3)\cdot (a-a^{\frac{2}{3}}b^{\frac{1}{5}})}{\small .}\)
Разложим выражения в числителе и знаменателе на множители:
- \(\displaystyle a^{\frac{2}{3}}-2a^{\frac{1}{3}}b^{\frac{1}{5}}+b^{\frac{2}{5}}=a^{\frac{2}{3}}-2\cdot a^{\frac{1}{3}}\cdot b^{\frac{1}{5}}+b^{\frac{2}{5}}=(a^{\frac{1}{3}}-b^{\frac{1}{5}})^2\small,\)
- \(\displaystyle a-a^{\frac{2}{3}}b^{\frac{1}{5}}=a^{\frac{2}{3}}(a^{\frac{1}{3}}-b^{\frac{1}{5}}){\small .}\)
Подставляя, получаем:
\(\displaystyle \frac{(a^{\frac{2}{3}}-2a^{\frac{1}{3}}b^{\frac{1}{5}}+b^{\frac{2}{5}})\cdot a^{\frac{2}{3}}}{(2b^{\frac{1}{5}}+3)\cdot (a-a^{\frac{2}{3}}b^{\frac{1}{5}})}= \frac{(a^{\frac{1}{3}}-b^{\frac{1}{5}})^2 \cdot a^{\frac{2}{3}}}{{(2b^{\frac{1}{5}}+3)}\cdot a^{\frac{2}{3}}(a^{\frac{1}{3}}-b^{\frac{1}{5}})} =\)
\(\displaystyle = \frac{a^{\frac{2}{3}}(a^{\frac{1}{3}}-b^{\frac{1}{5}})^2 }{{a^{\frac{2}{3}}(2b^{\frac{1}{5}}+3)}(a^{\frac{1}{3}}-b^{\frac{1}{5}})} {\small .}\)
\(\displaystyle \frac{a^{\frac{2}{3}}(a^{\frac{1}{3}}-b^{\frac{1}{5}})^2}{{a^{\frac{2}{3}}(2b^{\frac{1}{5}}+3)}(a^{\frac{1}{3}}-b^{\frac{1}{5}})} = \frac{a^{\frac{1}{3}}-b^{\frac{1}{5}}}{2b^{\frac{1}{5}}+3}\small. \)
Ответ: \(\displaystyle \frac{a^{\frac{1}{3}}-b^{\frac{1}{5}}}{2b^{\frac{1}{5}}+3} \small.\)