گام به گام تمرین صفحه 57 درس 3 ریاضی (1) (توان های گویا و عبارت های جبری)

1)

الف)

\({\left( {0/5} \right)^2}\;\;\) \(>\) \(\;\;{\left( {0/5} \right)^3}\)  ,  \(\sqrt {0/5} \;\;\) \(<\) \(\;\;\sqrt[3]{{0/5}}\)

\({4^2}\;\;\) \(<\) \(\;\;{4^3}\)  ,  \(\sqrt 4 \;\;\) \(>\) \(\;\;\sqrt[3]{4}\)

ب)

\({a^2}\;\;\) \(>\) \(\;\;{a^3}\)  ,  \(\sqrt a \;\;\) \(<\) \(\;\;\sqrt[3]{a}\)

پ)

\({a^2}\;\;\) \(<\) \(\;\;{a^3}\)  ,  \(\sqrt a \;\;\) \(>\) \(\;\;\sqrt[3]{a}\)

2)

الف)

\({\left( { - 0/5} \right)^2}\;\;\)\(>\) \(\;\;{\left( { - 0/5} \right)^3} \)   ,  \({\left( { - 2} \right)^2}\;\;\) \(>\) \(\;\;{\left( { - 2} \right)^3}\)

\({\left( { - 0/5} \right)^3}\;\;\) \(<\) \(\;\;{\left( { - 0/5} \right)^5}\)  ,  \({\left( { - 2} \right)^3}\;\;\) \(>\) \(\;\;{\left( { - 2} \right)^5}\)

\({\left( { - 0/5} \right)^4}\;\;\) \(<\) \(\;\;{\left( { - 0/5} \right)^2}\)  ,  \({\left( { - 2} \right)^4}\;\;\) \(>\) \(\;\;{\left( { - 2} \right)^2}\)

3)

\(b = \sqrt[n]{a} \Rightarrow {b^n} = {\left( {\sqrt[n]{a}} \right)^n} = a\)

4)

این تساوی همیشه برقرار نیست:

\(\begin{array}{l}\left. \begin{array}{l}\sqrt[3]{1} + \sqrt[3]{8} = 1 + 2 = 3\\\sqrt[3]{{1 + 8}} = \sqrt[3]{9} \simeq 2/08\end{array} \right\} \Rightarrow \sqrt[3]{1} + \sqrt[3]{8} \ne \sqrt[3]{{1 + 8}}\\\\\left. \begin{array}{l}\sqrt[4]{1} + \sqrt[4]{{81}} = 1 + 3 = 4\\\sqrt[4]{{1 + 81}} = \sqrt[4]{{82}} \simeq 3\end{array} \right\} \Rightarrow \sqrt[4]{1} + \sqrt[4]{{81}} \ne \sqrt[4]{{1 + 81}}\\\\\left. \begin{array}{l}\sqrt[5]{1} + \sqrt[5]{{32}} = 1 + 2 = 3\\\sqrt[5]{{1 + 32}} = \sqrt[5]{{33}} \simeq 3/2\end{array} \right\} \Rightarrow \sqrt[5]{1} + \sqrt[5]{{32}} \ne \sqrt[5]{{1 + 32}}\end{array}\)

5)

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

\(\sqrt[7]{{\frac{1}{{128}}}} =\) \(\sqrt[7]{{{{\left( {\frac{1}{2}} \right)}^7}}} = \frac{1}{2}\)

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

6)

الف)

\(\begin{array}{l}\left. \begin{array}{l}a = 8\\b = 27\\n = 3\end{array} \right\}\\\\ \Rightarrow \left\{ \begin{array}{l}\sqrt[3]{{\frac{8}{{27}}}} = \sqrt[3]{{{{\left( {\frac{2}{3}} \right)}^3}}} = \frac{2}{3}\\\frac{{\sqrt[3]{8}}}{{\sqrt[3]{{27}}}} = \frac{2}{3}\end{array} \right.\\\\ \Rightarrow \sqrt[3]{{\frac{8}{{27}}}} = \frac{{\sqrt[3]{8}}}{{\sqrt[3]{{27}}}}\end{array}\)

ب)

\(\begin{array}{l}\left. \begin{array}{l}a =  - \;16\\b =  - \;81\\n = 4\end{array} \right\}\\\\ \Rightarrow \left\{ \begin{array}{l}\sqrt[4]{{\frac{{ - \;16}}{{ - \;81}}}} = \sqrt[4]{{{{\left( {\frac{{ - 2}}{{ - 3}}} \right)}^4}}} = \frac{2}{3}\\\frac{{\sqrt[4]{{ - \;16}}}}{{\sqrt[4]{{ - \;81}}}} = \frac{ \otimes }{ \otimes }\end{array} \right.\\\\ \Rightarrow \sqrt[4]{{\frac{{ - \;16}}{{ - \;81}}}} = \frac{{\sqrt[4]{{ - \;16}}}}{{\sqrt[4]{{ - \;81}}}}\end{array} \)