گام به گام تمرین صفحه 61 درس 3 ریاضی (1) (توان های گویا و عبارت های جبری)

1)

\({16^{\frac{1}{2}}} =\) \(\sqrt[2]{{16}} = 4\)

\({5^{\frac{1}{2}}} =\) \(\sqrt[2]{5} = 5\)

\({4^{\frac{3}{7}}} =\) \(\sqrt[7]{{{4^3}}}\)

\({3^{\frac{1}{3}}} \times {3^{\frac{2}{3}}} =\) \(\sqrt[3]{3} \times \sqrt[3]{{{3^2}}} = \sqrt[3]{{{3^3}}} = 3\)

\({\left( {{4^{\frac{1}{2}}}} \right)^{\frac{2}{3}}} =\) \({4^{\frac{1}{3}}} = \sqrt[3]{4}\)

\({4^{\frac{2}{3}}} =\) \(\sqrt[3]{{{4^2}}}\)

\({32^{ - \frac{1}{5}}} =\) \(\frac{1}{{{{32}^{\frac{1}{5}}}}} = \frac{1}{{\sqrt[5]{{32}}}} = \frac{1}{2}\)

\({32^{\frac{2}{5}}} =\) \({\left( {\sqrt[5]{{32}}} \right)^2} = {2^2} = 4\)

\({125^{ - \frac{2}{3}}} =\) \(\frac{1}{{{{\left( {\sqrt[3]{{125}}} \right)}^2}}} = \frac{1}{{25}}\)

2)

بله؛ همیشه این تساوی برقرار است:

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

3)

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

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

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

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

نتیجه می گیریم که:

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

4)

\(\sqrt[3]{{\sqrt 5 }} =\) \(\sqrt[{3 \times 2}]{5} = \sqrt[6]{5}\)

\(\sqrt {\sqrt[3]{{64}}}  =\) \(\sqrt[{2 \times 3}]{{64}} = \sqrt[6]{{64}} = 2\)

\(\sqrt {\sqrt {81} }  =\) \(\sqrt[{2 \times 2}]{{81}} = \sqrt[4]{{81}} = 3\)