جواب تمرین صفحه 61 درس 3 ریاضی دهم (توان های گویا و عبارت های جبری)

\(\begin{array}{l}{16^{\frac{1}{2}}} = \sqrt[2]{{16}} = 4\\\\{5^{\frac{1}{2}}} = \sqrt[2]{5} = 5\\\\{4^{\frac{3}{7}}} = \sqrt[7]{{{4^3}}}\\\\{3^{\frac{1}{3}}} \times {3^{\frac{2}{3}}} = \sqrt[3]{3} \times \sqrt[3]{{{3^2}}} = \sqrt[3]{{{3^3}}} = 3\\\\{\left( {{4^{\frac{1}{2}}}} \right)^{\frac{2}{3}}} = {4^{\frac{1}{3}}} = \sqrt[3]{4}\\\\{4^{\frac{2}{3}}} = \sqrt[3]{{{4^2}}}\\\\{32^{ - \frac{1}{5}}} = \frac{1}{{{{32}^{\frac{1}{5}}}}} = \frac{1}{{\sqrt[5]{{32}}}} = \frac{1}{2}\\\\{32^{\frac{2}{5}}} = {\left( {\sqrt[5]{{32}}} \right)^2} = {2^2} = 4\\\\{125^{ - \frac{2}{3}}} = \frac{1}{{{{\left( {\sqrt[3]{{125}}} \right)}^2}}} = \frac{1}{{25}}\end{array}\)

بله؛ همیشه این تساوی برقرار است:

\(\begin{array}{l}\left\{ \begin{array}{l}\sqrt[6]{{{2^3}}} = {2^{\frac{3}{6}}} = {2^{\frac{1}{2}}} = \sqrt 2 \\\\\sqrt[4]{{{2^2}}} = {2^{\frac{2}{4}}} = {2^{\frac{1}{2}}} = \sqrt 2 \end{array} \right.\\\\ \Rightarrow \sqrt[6]{{{2^3}}} = \sqrt[4]{{{2^2}}} = \sqrt 2 \end{array}\)

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{l}}{a = 64}\\{r = \frac{1}{2}}\\{s = \frac{1}{3}}\end{array}} \right.\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\frac{{{a^r}}}{{{a^s}}} = \frac{{{{64}^{\frac{1}{2}}}}}{{{{64}^{\frac{1}{3}}}}} = \frac{{\sqrt {64} }}{{\sqrt[3]{{64}}}} = \frac{8}{4} = 2}\\{{a^{r - s}} = {{\left( {64} \right)}^{\frac{1}{2} - \frac{1}{3}}} = {{64}^{\frac{1}{6}}} = 2}\end{array}} \right.\\\\ \Rightarrow \frac{{{a^r}}}{{{a^s}}} = {a^{r - s}}\end{array}\)

نمونه اول:

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{l}}{a = 27}\\{r = \frac{2}{3}}\\{s = \frac{1}{3}}\end{array}} \right.\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\frac{{{a^r}}}{{{a^s}}} = \frac{{{{27}^{\frac{2}{3}}}}}{{{{27}^{\frac{1}{3}}}}} = \frac{{{{\left( {\sqrt[3]{{27}}} \right)}^2}}}{{\sqrt[3]{{27}}}} = \frac{{{3^2}}}{3} = 3}\\{{a^{r - s}} = {{\left( {27} \right)}^{\frac{2}{3} - \frac{1}{3}}} = {{27}^{\frac{1}{3}}} = 3}\end{array}} \right.\\\\ \Rightarrow \frac{{{a^r}}}{{{a^s}}} = {a^{r - s}}\end{array}\)

نمونه دوم:

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{l}}{a = 256}\\{r = \frac{1}{2}}\\{s = \frac{1}{4}}\end{array}} \right.\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\frac{{{a^r}}}{{{a^s}}} = \frac{{{{256}^{\frac{1}{2}}}}}{{{{256}^{\frac{1}{4}}}}} = \frac{{\sqrt[2]{{256}}}}{{\sqrt[4]{{256}}}} = \frac{{16}}{4} = 4}\\{{a^{r - s}} = {{\left( {256} \right)}^{\frac{1}{2} - \frac{1}{4}}} = {{256}^{\frac{1}{4}}} = 4}\end{array}} \right.\\\\ \Rightarrow \frac{{{a^r}}}{{{a^s}}} = {a^{r - s}}\end{array}\)

نمونه سوم:

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{l}}{a = 1024}\\{r = \frac{1}{2}}\\{s = \frac{1}{5}}\end{array}} \right.\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\frac{{{a^r}}}{{{a^s}}} = \frac{{{{1024}^{\frac{1}{2}}}}}{{{{1024}^{\frac{1}{5}}}}} = \frac{{\sqrt {1024} }}{{\sqrt[5]{{1024}}}} = \frac{{32}}{4} = 8}\\{{a^{r - s}} = {{\left( {1024} \right)}^{\frac{1}{2} - \frac{1}{5}}} = {{1024}^{\frac{3}{{10}}}} = {{\left( {\sqrt[{10}]{{1024}}} \right)}^3} = {2^3} = 8}\end{array}} \right.\\\\ \Rightarrow \frac{{{a^r}}}{{{a^s}}} = {a^{r - s}}\end{array}\)

از نتایج به دست آمده نتیجه می گیریم که:

\(\frac{{{a^r}}}{{{a^s}}} = {a^{r - s}}\)

\(\begin{array}{l}\sqrt[3]{{\sqrt 5 }} = \sqrt[{3 \times 2}]{5} = \sqrt[6]{5}\\\\\sqrt {\sqrt[3]{{64}}} = \sqrt[{2 \times 3}]{{64}} = \sqrt[6]{{64}} = 2\\\\\sqrt {\sqrt {81} } = \sqrt[{2 \times 2}]{{81}} = \sqrt[4]{{81}} = 3\end{array}\)