جواب تمرین صفحه 71 درس 4 ریاضی نهم (توان و ریشه)

\(\begin{array}{l}\frac{{{\rm{49}}}}{{{\rm{16}}}} \Rightarrow \frac{{\rm{7}}}{{\rm{4}}}, - \frac{{\rm{7}}}{{\rm{4}}}\\\\\frac{{\rm{1}}}{{{\rm{81}}}} \Rightarrow \frac{{\rm{1}}}{{\rm{9}}}{\rm{,}} - \frac{{\rm{1}}}{{\rm{9}}}{\rm{\;\;\;\;\;\;\;\;\;\;}}\\\\{\rm{15}} \Rightarrow \sqrt {{\rm{15}}} \,,\, - \sqrt {{\rm{15}}} \\\\13 \Rightarrow 2\sqrt 3 \,,\, - 2\sqrt 3 \\\\144 \Rightarrow 12\,,\, - 12\\\\18 \Rightarrow 3\sqrt 2 \,,\, - 3\sqrt 2 \end{array}\)

\(\begin{array}{l}216 \Rightarrow 6\\\\{7^3} \Rightarrow 7\\\\ - 5 \Rightarrow \sqrt[3]{{ - 5}}\\\\ - \frac{1}{{216}} \Rightarrow - \frac{1}{6}\\\\10 \to \sqrt[3]{{10}}\end{array}\)

عبارات صحیح از قرار زیر هستند:

\(\begin{array}{l}\sqrt[3]{{{{\left( { - 1} \right)}^3}}} = - 1\\\\\sqrt {{{\left( { - 5} \right)}^2}} = \left| { - 5} \right| = 5\\\\\sqrt[3]{{{{\left( { - 5} \right)}^3}}} = - 5\\\\ - \sqrt[3]{{\frac{{49}}{{256}}}} = - \frac{7}{{16}}\\\\\sqrt {\frac{1}{{44}}} = 1/2\\\\\sqrt[3]{{ - 64}} = - 4\end{array}\)

\(\begin{array}{l}\sqrt[3]{a} < \sqrt 4 \\\\ \Rightarrow \sqrt[3]{a} < 2\;\\\\ \Rightarrow \sqrt[3]{a} < \sqrt[3]{8}\\\\ \Rightarrow a < 8\end{array}\)

\(\begin{array}{l}\sqrt {{{\left( { - x} \right)}^2}} = \sqrt {{{\left( x \right)}^2}} = \left| x \right|\;\\\\\\\sqrt {{{\left( { - x} \right)}^2}} = x\end{array}\)

در صورتی که \(\left| {\rm{x}} \right| = x\;\)  است که \(x \ge 0\)  باشد.

\(\begin{array}{l}\sqrt {{{\left( { - 2} \right)}^2}} = 2\\\\\\\sqrt {{{\left( { - \left( { - 2} \right)} \right)}^2}} \ne - 2\;{\rm{\;}}\end{array}\)

\(\begin{array}{l}{\rm{\;}}S = 96{a^2}\\\\ \Rightarrow \;S = 96{a^2} \div 6 = 16{a^2}\;{\rm{\;}}\end{array}\)

نمی دانیم که مقدار a مثبت است یا منفی

\(\begin{array}{l}S = \sqrt {16{a^2}} = \left| {4a} \right|\;{\rm{\;}}\\\\ \Rightarrow V = {\left( {\left| {4a} \right|} \right)^3} = 64{a^2}\left| a \right|\\\\\mathop \Rightarrow \limits^{a > 0} \,\,\,\,\,V = 64{a^3}\;{\rm{\;}}\end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}x > 0\; \Rightarrow \sqrt {{x^2}} = \left| x \right| = x\\\\y < 0 \Rightarrow \sqrt {{y^2}} = \left| y \right| = - y\end{array} \right.\\\\ \Rightarrow \sqrt {{x^2}} - \sqrt {{y^2}} = \left| x \right| - \left| y \right| = x - \left( { - y} \right) = x + y\;\end{array}\)

\(\begin{array}{l}\sqrt {150} = \sqrt {2 \times 3 \times {5^2}} = \\\\\sqrt {2 \times 3} \times \sqrt {{5^2}} = \\\\\sqrt 6 \times 5 = 5\sqrt 6 \end{array}\)

\(\begin{array}{l}\sqrt {80} = \sqrt {{2^4} \times {5^2}} = \\\\\sqrt 5 \times \sqrt {{2^4}} = 4\sqrt 5 \end{array}\)

\(\begin{array}{l}\sqrt {24} = \sqrt {3 \times {2^3}} = \\\\\sqrt {3 \times {2^3} \times 2} \times \sqrt {{2^2}} = \\\\\sqrt {{2^2}} \times 5\sqrt 6 = 2\sqrt 6 \end{array}\)

\(\begin{array}{l}\sqrt[3]{{{{125}^2}}} = \sqrt[3]{{{{\left( {{5^3}} \right)}^2}}} = \\\\\sqrt[3]{{{5^6}}} = \sqrt[3]{{{{\left( {{5^2}} \right)}^3}}} = \\\\{5^2} = 25\end{array}\)

\(\begin{array}{l}{\left( {\sqrt[3]{{ - 2}}} \right)^3} = \left( {\sqrt[3]{{ - 2}} \times \sqrt[3]{{ - 2}} \times \sqrt[3]{{ - 2}}} \right) = \\\\\sqrt[3]{{\left( { - 2} \right)\left( { - 2} \right)\left( { - 2} \right)}} = \sqrt[3]{{{{\left( { - 2} \right)}^3}}} = - 2\end{array}\)

\(\begin{array}{l}\sqrt[3]{{ - 4}} = \sqrt[3]{{\left( { - 1} \right) \times 4}} = \\\\\sqrt[3]{{ - 1}} \times \sqrt[3]{4} = - 1 \times \sqrt[3]{4} = - \sqrt[3]{4}\end{array}\)

\(\begin{array}{l}2\sqrt[3]{{16}} \times 3\sqrt[3]{4} = 6 \times \sqrt[3]{{16 \times 4}} = \\\\6\sqrt[3]{{64}} = 6 \times 4 = 24\end{array}\)

\(\begin{array}{l}\frac{{\sqrt 8 \times \sqrt 5 }}{{\sqrt {10} }} = \sqrt {\frac{{8 \times 5}}{{10}}} = \\\\\sqrt {\frac{{40}}{{10}}} = \sqrt 4 = 2\end{array}\)

\(\begin{array}{l}\frac{{\sqrt[3]{{18}} \times \sqrt[3]{{60}}}}{{\sqrt[3]{5}}} = \sqrt[3]{{\frac{{18 \times 60}}{5}}} = \\\\\sqrt[3]{{216}} = 6\end{array}\)