Решение.

Применяем свойства степеней   [tex]\bf a^{n}\cdot a^{k}=a^{n+k}\ ,\ \ (a^{n})^{k}=a^{n\cdot k}[/tex]и формулы сокращённого умножения:

[tex]\bf a^2-b^2=(a-b)(a+b)\ ,\ \ (a+b)^2=a^2+2ab+b^2[/tex]  .

[tex]\displaystyle \bf 1)\ \ \frac{x+7x^{\frac{2}{5}}}{x^{\frac{3}{5}}+7}=\frac{x^{\frac{2}{5}}\cdot (x^{\frac{3}{5}}+7)}{x^{\frac{3}{5}}+7}=x^{\frac{2}{5}}\\\\\\2)\ \ \frac{a^{\frac{1}{3}}-b^{\frac{1}{3}}}{a^{\frac{1}{6}}-b^{\frac{1}{6}}}=\frac{(a^{\frac{1}{6}})^2-(b^{\frac{1}{6}})^2}{a^{\frac{1}{6}}-b^{\frac{1}{6}}}=\frac{(a^{\frac{1}{6}}-b^{\frac{1}{6}})(a^{\frac{1}{6}}+b^{\frac{1}{6}})}{a^{\frac{1}{6}}-b^{\frac{1}{6}}}=a^{\frac{1}{6}}+b^{\frac{1}{6}}[/tex]

[tex]\bf \displaystyle 3)\ \ \frac{m^{\frac{1}{2}}n^{\frac{1}{4}}+3m^{\frac{1}{4}}n^{\frac{1}{2}}}{m^{\frac{1}{2}}+6m^{\frac{1}{4}}n^{\frac{1}{4}}+9n^{\frac{1}{2}}}=\frac{m^{\frac{1}{4}}n^{\frac{1}{4}}\cdot (m^{\frac{1}{4}}+3n^{\frac{1}{4}})}{(m^{\frac{1}{4}}+3n^{\frac{1}{4}})^2}=\frac{m^{\frac{1}{4}}n^{\frac{1}{4}}}{m^{\frac{1}{4}}+3n^{\frac{1}{4}}}[/tex]