جواب تمرین صفحه 57 درس 3 ریاضی دهم (توان های گویا و عبارت های جبری)

الف

\(\begin{array}{l}{\left( {0/5} \right)^2}\;\; > \;\;{\left( {0/5} \right)^3}\\\\\sqrt {0/5} \;\; < \;\;\sqrt[3]{{0/5}}\\\\{4^2}\;\; < \;\;{4^3}\\\\\sqrt 4 \;\; > \;\;\sqrt[3]{4}\end{array}\)

ب

\(\begin{array}{l}{a^2}\;\; > \;\;{a^3}\\\\\sqrt a \;\; < \;\;\sqrt[3]{a}\end{array}\)

پ

\(\begin{array}{l}{a^2}\;\; < \;\;{a^3}\\\\\sqrt a \;\; > \;\;\sqrt[3]{a}\end{array}\)

\(\begin{array}{l}{\left( { - 0/5} \right)^2}\;\; > \;\;{\left( { - 0/5} \right)^3}\\\\{\left( { - 2} \right)^2}\;\; > \;\;{\left( { - 2} \right)^3}\\\\{\left( { - 0/5} \right)^3}\;\; < \;\;{\left( { - 0/5} \right)^5}\\\\{\left( { - 2} \right)^3}\;\; > \;\;{\left( { - 2} \right)^5}\\\\{\left( { - 0/5} \right)^4}\;\; < \;\;{\left( { - 0/5} \right)^2}\\\\{\left( { - 2} \right)^4}\;\; > \;\;{\left( { - 2} \right)^2}\end{array}\)

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

این تساوی همیشه برقرار نیست:

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

\(\begin{array}{l}\sqrt[5]{{{2^{ - 5}}}} = \sqrt[5]{{{{\left( {\frac{1}{2}} \right)}^5}}} = \frac{1}{2}\\\\\sqrt[7]{{\frac{1}{{128}}}} = \sqrt[7]{{{{\left( {\frac{1}{2}} \right)}^7}}} = \frac{1}{2}\\\\\sqrt[4]{{{3^{ - 4}}}} = \sqrt[4]{{{{\left( {\frac{1}{3}} \right)}^4}}} = \frac{1}{3}\end{array}\)

الف

\(\begin{array}{*{20}{l}}{\left. {\begin{array}{*{20}{l}}{a = 8}\\{b = 27}\\{n = 3}\end{array}} \right\}}\\{}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\sqrt[3]{{\frac{8}{{27}}}} = \sqrt[3]{{{{\left( {\frac{2}{3}} \right)}^3}}} = \frac{2}{3}}\\{\frac{{\sqrt[3]{8}}}{{\sqrt[3]{{27}}}} = \frac{2}{3}}\end{array}} \right.}\\{}\\{ \Rightarrow \sqrt[3]{{\frac{8}{{27}}}} = \frac{{\sqrt[3]{8}}}{{\sqrt[3]{{27}}}}}\end{array}\)

ب

\(\begin{array}{*{20}{l}}{\left. {\begin{array}{*{20}{l}}{a = - \;16}\\{b = - \;81}\\{n = 4}\end{array}} \right\}}\\{}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\sqrt[4]{{\frac{{ - \;16}}{{ - \;81}}}} = \sqrt[4]{{{{\left( {\frac{{ - 2}}{{ - 3}}} \right)}^4}}} = \frac{2}{3}}\\{\frac{{\sqrt[4]{{ - \;16}}}}{{\sqrt[4]{{ - \;81}}}} = \frac{ \otimes }{ \otimes }}\end{array}} \right.}\\{}\\{ \Rightarrow \sqrt[4]{{\frac{{ - \;16}}{{ - \;81}}}} = \frac{{\sqrt[4]{{ - \;16}}}}{{\sqrt[4]{{ - \;81}}}}}\end{array}\)