Упростите выражение:
Определим порядок действий:
\(\displaystyle \color{Blue}{\left(\frac{3n^{\frac{2}{5}}}{n^{\frac{2}{5}}-5} \overset {\color {red}{\bf1}}{-} \frac{14n^{\frac{2}{5}}}{n^{\frac{4}{5}}-10n^{\frac{2}{5}}+25}\right) }\overset {\color {red}{\bf2}}{ \overset{\phantom {}}{\,\cdot \,}} \color{green}{\frac{n^{\frac{4}{5}}-25}{3n^{\frac{2}{5}}-29}} \overset {\color {red}{\bf3}}{-} \color{purple}{\frac{5(n^{\frac{2}{5}}+5)}{n^{\frac{2}{5}}-5}} {\small. }\)
\(\displaystyle \color{Blue}{\left(\frac{3n^{\frac{2}{5}}}{n^{\frac{2}{5}}-5} {-} \frac{14n^{\frac{2}{5}}}{n^{\frac{4}{5}}-10n^{\frac{2}{5}}+25}\right) =\frac{n^{\frac{2}{5}}(3n^{\frac{2}{5}}-29)}{(n^{\frac{2}{5}}-5)^2}{\small .}}\)
В знаменателе второй дроби находится квадрат разности:
\(\displaystyle n^{\frac{4}{5}}-10n^{\frac{2}{5}}+25=(n^{\frac{2}{5}})^2-2\cdot 5\cdot n^{\frac{2}{5}}+5^2=(n^{\frac{2}{5}}-5)^2{\small. }\)
Тогда общий знаменатель дробей \(\displaystyle \frac{3n^{\frac{2}{5}}}{n^{\frac{2}{5}}-5}\) и \(\displaystyle \frac{14n^{\frac{2}{5}}}{n^{\frac{4}{5}}-10n^{\frac{2}{5}}+25}\) равен \(\displaystyle (n^{\frac{2}{5}}-5)^2{\small. }\)
Приведём дроби к общему знаменателю:
\(\displaystyle \frac{3n^{\frac{2}{5}}}{n^{\frac{2}{5}}-5} - \frac{14n^{\frac{2}{5}}}{n^{\frac{4}{5}}-10n^{\frac{2}{5}}+25}=\frac{3n^{\frac{2}{5}}}{n^{\frac{2}{5}}-5} - \frac{14n^{\frac{2}{5}}}{(n^{\frac{2}{5}}-5)^2}=\frac{3n^{\frac{2}{5}}(n^{\frac{2}{5}}-5)-14n^{\frac{2}{5}}}{(n^{\frac{2}{5}}-5)^2}{\small. }\)
Раскроем скобки в числителе и приведём подобные:
\(\displaystyle \frac{3n^{\frac{2}{5}}(n^{\frac{2}{5}}-5)-14n^{\frac{2}{5}}}{(n^{\frac{2}{5}}-5)^2}=\frac{3n^{\frac{4}{5}}-15n^{\frac{2}{5}}-14n^{\frac{2}{5}}}{(n^{\frac{2}{5}}-5)^2}=\frac{3n^{\frac{4}{5}}-29n^{\frac{2}{5}}}{(n^{\frac{2}{5}}-5)^2}{\small. }\)
Вынесем общий множитель в числителе:
\(\displaystyle \frac{3n^{\frac{4}{5}}-29n^{\frac{2}{5}}}{(n^{\frac{2}{5}}-5)^2}=\frac{n^{\frac{2}{5}}(3n^{\frac{2}{5}}-29)}{(n^{\frac{2}{5}}-5)^2}{\small .}\)
\(\displaystyle \color{Blue}{\frac{n^{\frac{2}{5}}(3n^{\frac{2}{5}}-29)}{(n^{\frac{2}{5}}-5)^2}} \cdot \color{green}{\frac{n^{\frac{4}{5}}-25}{3n^{\frac{2}{5}}-29}} = \frac{n^{\frac{2}{5}}(3n^{\frac{2}{5}}-29)(n^{\frac{4}{5}}-25)}{(n^{\frac{2}{5}}-5)^2(3n^{\frac{2}{5}}-29)} {\small. }\)
Воспользуемся формулой разности квадратов для выражения \(\displaystyle n^{\frac{4}{5}}-25\) и сократим дробь:
\(\displaystyle \frac{n^{\frac{2}{5}}(3n^{\frac{2}{5}}-29)(n^{\frac{4}{5}}-25)}{(n^{\frac{2}{5}}-5)^2(3n^{\frac{2}{5}}-29)} = \frac{n^{\frac{2}{5}}(3n^{\frac{2}{5}}-29)(n^{\frac{2}{5}}-5)(n^{\frac{2}{5}}+5)}{(n^{\frac{2}{5}}-5)^2(3n^{\frac{2}{5}}-29)}= \color{green}{\frac{n^{\frac{2}{5}}(n^{\frac{2}{5}}+5)}{n^{\frac{2}{5}}-5}}{\small. }\)
\(\displaystyle \color{green}{\frac{n^{\frac{2}{5}}(n^{\frac{2}{5}}+5)}{n^{\frac{2}{5}}-5}} - \color{purple}{\frac{5(n^{\frac{2}{5}}+5)}{n^{\frac{2}{5}}-5}}=\frac{n^{\frac{2}{5}}(n^{\frac{2}{5}}+5)-5(n^{\frac{2}{5}}+5)}{n^{\frac{2}{5}}-5}=\frac{(n^{\frac{2}{5}}-5)(n^{\frac{2}{5}}+5)}{n^{\frac{2}{5}}-5}=n^{\frac{2}{5}}+5{\small. }\)
Ответ: \(\displaystyle n^{\frac{2}{5}}+5{\small. }\)