Potęga o wykładniku wymiernym 1 przez n

Potęga o wykładniku wymiernym 1 przez n

• \left.\mleft.121\mright.\right.^{\mleft.\textcolor{#E94D00}{\frac{1}{2}}\mright.}=\sqrt[{\textcolor{#E94D00}{2}}]{121}=11,\left.\right. 12112  =  2√121  =  11,  \left.\mleft.8\mright.\right.^{\mleft.\textcolor{#E94D00}{\frac{1}{3}}\mright.}=\sqrt[{\textcolor{#E94D00}{3}}]{8}=2,\left.\right. 813  =  3√8  =  2,  \left.\mleft.243\mright.\right.^{\mleft.\textcolor{#E94D00}{\frac{1}{5}}\mright.}=\sqrt[{\textcolor{#E94D00}{5}}]{243}=3 24315  =  5√243  =  3
• \left.\mleft.11\mright.\right.^{\mleft.\textcolor{#E94D00}{\frac{1}{2}}\mright.}=\sqrt[{\textcolor{#E94D00}{2}}]{11},\left.\right. 1112  =  2√11,  \left.\mleft.7\mright.\right.^{\mleft.\textcolor{#E94D00}{\frac{1}{3}}\mright.}=\sqrt[{\textcolor{#E94D00}{3}}]{7},\left.\right. 713  =  3√7,  \left.\mleft.5\mright.\right.^{\mleft.\textcolor{#E94D00}{\frac{1}{n}}\mright.}=\sqrt[{\textcolor{#E94D00}{n}}]{5} 51n  =  n√5
• \left.\mleft.a\mright.\right.^{\mleft.\textcolor{#E94D00}{\frac{1}{n}}\mright.}=\sqrt[{\textcolor{#E94D00}{n}}]{a},\left.\right. a1n  =  n√a,  gdy a\left.\right.{{\fontfamily{ams}\geqslant }}\left.\right.0 a ⩾ 0
• Wyznaczanie potęgi o wykładniku \frac{1}{n} 1n to obliczenie pierwiastka stopnia n n .