Ответ:
Применяем свойства степеней:
[tex]\boldsymbol{(a^{n})^{k}=a^{nk}\ \ ,\ \ a^{n}\cdot a^{k}=a^{n+k}\ \ ,\ \ \dfrac{a^{n}}{a^{k}}=a^{n-k}\ \ ,\ \ \sqrt[2n-1]{\bf a^{2n-1}}=a}[/tex] .
[tex]\boldsymbol{2)\ \ \left(\Big(\sqrt[3]{\bf 5}\Big)^{\sqrt3}\right)^{\sqrt3}=\Big(\sqrt[3]{\bf 5}\Big)^{\sqrt3\cdot \sqrt3}=\Big(\sqrt[3]{\bf 5}\Big)^{3}=\sqrt[3]{5^3}=5}[/tex]
[tex]\boldsymbol{4)\ \ 6^{^{2\sqrt5+1}}\, :\, 36^{^{\sqrt5}}=6^{^{2\sqrt5}}\cdot 6^1\, :\, (6^2)^{^{\sqrt5}}=6^{^{2\sqrt5}}\cdot 6^1\, :\, 6^{^{2\sqrt5}}=\dfrac{6^{^{2\sqrt5}}\cdot 6}{6^{^{2\sqrt5}}}=6}[/tex]
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Ответ:
Применяем свойства степеней:
[tex]\boldsymbol{(a^{n})^{k}=a^{nk}\ \ ,\ \ a^{n}\cdot a^{k}=a^{n+k}\ \ ,\ \ \dfrac{a^{n}}{a^{k}}=a^{n-k}\ \ ,\ \ \sqrt[2n-1]{\bf a^{2n-1}}=a}[/tex] .
[tex]\boldsymbol{2)\ \ \left(\Big(\sqrt[3]{\bf 5}\Big)^{\sqrt3}\right)^{\sqrt3}=\Big(\sqrt[3]{\bf 5}\Big)^{\sqrt3\cdot \sqrt3}=\Big(\sqrt[3]{\bf 5}\Big)^{3}=\sqrt[3]{5^3}=5}[/tex]
[tex]\boldsymbol{4)\ \ 6^{^{2\sqrt5+1}}\, :\, 36^{^{\sqrt5}}=6^{^{2\sqrt5}}\cdot 6^1\, :\, (6^2)^{^{\sqrt5}}=6^{^{2\sqrt5}}\cdot 6^1\, :\, 6^{^{2\sqrt5}}=\dfrac{6^{^{2\sqrt5}}\cdot 6}{6^{^{2\sqrt5}}}=6}[/tex]