ГДЗ по алгебре 10 класс Алимов Ш. А., Колягин Ю. М., Ткачёва М. В. и др. упражнение - 78 стр. 33

$1) \frac{a^{{\displaystyle \frac{4}{3}}}\left(a^{—{\displaystyle \frac{1}{3}}}+a^{{\displaystyle \frac{2}{3}}}\right)}{a^{\frac{1}{4}}\left(a^{\frac{3}{4}}+a^{—\frac{1}{4}}\right)}.$

Раскроем скобки и воспользуемся свойством степеней

$a^n·a^m=a^{n+m}.$ $\frac{a^{{\displaystyle \frac{4}{3}}}\left(a^{—{\displaystyle \frac{1}{3}}}+a^{{\displaystyle \frac{2}{3}}}\right)}{a^{\frac{1}{4}}\left(a^{\frac{3}{4}}+a^{—\frac{1}{4}}\right)}=\frac{a^{{\displaystyle \frac{4}{3}}}·a^{—{\displaystyle \frac{1}{3}}}+a^{\frac{4}{3}}·a^{{\displaystyle \frac{2}{3}}}}{a^{\frac{1}{4}}·a^{\frac{3}{4}}+a^{\frac{1}{4}}·a^{—\frac{1}{4}}}=\frac{a^{{\displaystyle \frac{4}{3}+\left(—\frac{1}{3}\right)}}+a^{\frac{4}{3}+{\displaystyle \frac{2}{3}}}}{a^{\frac{1}{4}+{\displaystyle \frac{3}{4}}}+a^{\frac{1}{4}+\left(—{\displaystyle \frac{1}{4}}\right)}}=\frac{a^{{\displaystyle 1}}+a^2}{a^1+a^0}=\frac{a+a^2}{a+1}=\frac{a\left(1+a\right)}{1+a}=a.$ $2) \frac{b^{{\displaystyle \frac{1}{5}}}\left(\sqrt[5]{b^4}—\sqrt[5]{b^{—1}}\right)}{b^{\frac{2}{3}}\left(\sqrt[3]{b}—\sqrt[3]{b^{—2}}\right)}.$

Раскроем скобки и воспользуемся свойством корней

$\sqrt[n]{a^m}=a^{\frac{m}{n}}$

и свойством степеней

$a^n·a^m=a^{n+m}.$ $\frac{b^{{\displaystyle \frac{1}{5}}}\left(\sqrt[5]{b^4}—\sqrt[5]{b^{—1}}\right)}{b^{\frac{2}{3}}\left(\sqrt[3]{b}—\sqrt[3]{b^{—2}}\right)}=\frac{b^{{\displaystyle \frac{1}{5}}}\left(b^{\frac{4}{5}}—b^{—{\displaystyle \frac{1}{5}}}\right)}{b^{\frac{2}{3}}\left(b^{{\displaystyle \frac{1}{3}}}—b^{—{\displaystyle \frac{2}{3}}}\right)}=\frac{b^{{\displaystyle \frac{1}{5}}}·b^{\frac{4}{5}}—b^{\frac{1}{5}}·b^{—{\displaystyle \frac{1}{5}}}}{b^{\frac{2}{3}}·b^{{\displaystyle \frac{1}{3}}}—b^{\frac{2}{3}}·b^{—{\displaystyle \frac{2}{3}}}}=\frac{b^{{\displaystyle \frac{1}{5}+\frac{4}{5}}}—b^{\frac{1}{5}+\left(—{\displaystyle \frac{1}{5}}\right)}}{b^{\frac{2}{3}+{\displaystyle \frac{1}{3}}}—b^{\frac{2}{3}+\left(—{\displaystyle \frac{2}{3}}\right)}}=\frac{b^{{\displaystyle \frac{5}{5}}}—b^{\frac{1}{5}—{\displaystyle \frac{1}{5}}}}{b^{\frac{3}{3}}—b^{\frac{2}{3}—{\displaystyle \frac{2}{3}}}}=\frac{b^{{\displaystyle 1}}—b^0}{b^1—b^0}=\frac{b—1}{b—1}=1.$ $3) \frac{a^{{\displaystyle \frac{5}{3}}}{b}^{—1}—a^{—{\displaystyle \frac{1}{3}}}}{\sqrt[3]{a^2}}.$

Воспользуемся свойствами степеней

$a^n:a^m=a^{n—m}, a^{—n}=\frac{1}{a^n}.$ $\frac{a^{{\displaystyle \frac{5}{3}}}{b}^{—1}—a^{—{\displaystyle \frac{1}{3}}}}{a^{{\displaystyle \frac{2}{3}}}}=\frac{a^{{\displaystyle \frac{5}{3}}}{b}^{—1}}{a^{\frac{2}{3}}}—\frac{a^{—{\displaystyle \frac{1}{3}}}}{a^{\frac{2}{3}}}=a^{\frac{5}{3}—\frac{2}{3}}{b}^{—1}—a^{—\frac{1}{3}—\frac{2}{3}}=a^{\frac{3}{3}}·\frac{1}{b}—a^{—\frac{3}{3}}=\frac{a}{b}—a^{—1}=\frac{a}{b}—\frac{1}{a}=\frac{a^2}{ab}—\frac{b}{ab}=\frac{a^2—b}{ab}.$ $4) \frac{a^{{\displaystyle \frac{1}{3}}}\sqrt{b}+b^{{\displaystyle \frac{1}{3}}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}.$

Воспользуемся свойством корней

$\sqrt[n]{a^m}=a^{\frac{m}{n}}$

и свойством степеней

$a^n·b^n={\left(ab\right)}^n.$ $\frac{a^{{\displaystyle \frac{1}{3}}}\sqrt{b}+b^{{\displaystyle \frac{1}{3}}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}=\frac{a^{{\displaystyle \frac{1}{3}}}{b}^{{\displaystyle \frac{1}{2}}}+b^{{\displaystyle \frac{1}{3}}}{a}^{{\displaystyle \frac{1}{2}}}}{a^{{\displaystyle \frac{1}{6}}}+b^{{\displaystyle \frac{1}{6}}}}=\frac{a^{{\displaystyle \frac{2}{6}}}{b}^{{\displaystyle \frac{3}{6}}}+b^{{\displaystyle \frac{2}{6}}}{a}^{{\displaystyle \frac{3}{6}}}}{a^{{\displaystyle \frac{1}{6}}}+b^{{\displaystyle \frac{1}{6}}}}=\frac{a^{{\displaystyle \frac{2}{6}}}{b}^{{\displaystyle \frac{2}{6}}}\left(b^{{\displaystyle \frac{1}{6}}}+a^{{\displaystyle \frac{1}{6}}}\right)}{a^{{\displaystyle \frac{1}{6}}}+b^{{\displaystyle \frac{1}{6}}}}=a^{\frac{2}{6}}{b}^{\frac{2}{6}}=a^{\frac{1}{3}}{b}^{\frac{1}{3}}={\left(ab\right)}^{\frac{1}{3}}=\sqrt[3]{ab}.$