جواب تمرین صفحه 93 درس 3 ریاضی و آمار دوازدهم انسانی (الگوهای غیر خطی)

\(\left. {\begin{array}{*{20}{l}}{\sqrt 5 = {5^{\frac{1}{2}}}}\\{\sqrt[4]{{{5^2}}} = {5^{\frac{2}{4}}} = {5^{\frac{1}{2}}}}\\{\sqrt[6]{{{5^3}}} = {5^{\frac{3}{6}}} = {5^{\frac{1}{2}}}}\end{array}} \right\} \Rightarrow \sqrt 5 = \sqrt[4]{{{5^2}}} = \sqrt[6]{{{5^3}}}\)

\({3^{\frac{1}{3}}} \times {3^{\frac{1}{4}}} = {3^{\frac{1}{3}\: + \:\frac{1}{4}}} = {3^{\frac{7}{{12}}}} = \sqrt[{12}]{{{3^7}}}\)

\({5^{\frac{1}{4}}} \times {5^{\left( { - \frac{1}{4}} \right)}} = {5^{\frac{1}{4}\: + \:( - \:\frac{1}{4})}} = {5^0} = 1 \)

\({8^{\frac{1}{2}}} \times {2^{\frac{1}{2}}} = {(8 \times 2)^{\frac{1}{2}}} = {16^{\frac{1}{2}}} = \sqrt {16} = 4\)

\({\left( {{2^6}} \right)^{\frac{1}{3}}} = {2^{6 \times \frac{1}{3}}} = {2^2} = 4\)

\({\left( {\frac{{{3^4}}}{{{2^6}}}} \right)^{\frac{1}{2}}} = \frac{{{3^{4 \times \frac{1}{2}}}}}{{{2^{6 \times \frac{1}{2}}}}} = \frac{{{3^2}}}{{{2^3}}} = \frac{9}{8}\)

\({\left( {\frac{{{a^{ - \frac{1}{2}}}}}{{{a^{ - \frac{1}{4}}}}}} \right)^{ - 4}} = \frac{{{a^{( - \frac{1}{2}) \times ( - 4)}}}}{{{a^{( - \frac{1}{4}) \times ( - 4)}}}} = \frac{{{a^2}}}{a} = 4\)

\({3^{0/26}} \times {3^{0/74}} = {3^{0/26 + 0/74}} = {3^1} = 3\)

\(\begin{array}{l}{\left( {{m^{\frac{3}{4}}}\;.\;{n^{\frac{1}{2}}}} \right)^2}{\left( {{m^2}\;.\;{n^3}} \right)^{\frac{1}{2}}} = {m^{\frac{3}{4} \times 2 + 2 \times \frac{1}{2}}}.{n^{\frac{1}{2} \times 2 + 3 \times \frac{1}{2}}}\\\\ = {m^{\frac{5}{2}}}.{n^{\frac{7}{2}}} = \sqrt {{m^5}.{n^7}} \end{array}\)

\(\begin{array}{l}{8^x} \times {9^5} = {72^5}\\\\ \Rightarrow {8^x} \times {9^5} = {(8 \times 9)^5} \Rightarrow {8^x} \times {9^5} = {8^5} \times {9^5}\\\\ \Rightarrow {8^x} = {8^5} \Rightarrow x = 5\end{array}\)

\(\begin{array}{l}{\left( {0/36} \right)^4} \times {\left( {0/36} \right)^x} \times {\left( {0/36} \right)^{ - 6}} = {\left( {0/36} \right)^7}\\\\ \Rightarrow {(0/36)^{4 + x - 6}} = {(0/36)^7}\\\\ \Rightarrow 4 + x - 6 = 7 \Rightarrow x = 9\end{array}\)

\(\begin{array}{l}{\left( {{3^x}} \right)^6} = \frac{1}{{{3^2}}}\\\\ \Rightarrow {3^{6x}} = {3^{ - 2}} \Rightarrow 6x = - 2 \Rightarrow x = - \frac{1}{3}\end{array}\)

\(\begin{array}{l}\frac{{{x^5} \times {{15}^3}}}{{{3^2} \times {3^5} \times 3}} = {5^8}\\\\ \Rightarrow {x^5} = \frac{{{5^8} \times {3^{2 + 5 + 1}}}}{{{{15}^3}}} = \frac{{{5^8} \times {3^8}}}{{{{15}^3}}}\\\\ = \frac{{{{15}^8}}}{{{{15}^3}}} = {15^5} \Rightarrow x = 15\end{array}\)

الف

\(\begin{array}{l}r = {\left( {\frac{{3v}}{{4\pi }}} \right)^{\frac{1}{3}}}\quad \mathop = \limits^{(v = \frac{4}{3}\pi {r^3})} \quad {\left( {\frac{{3(\frac{4}{3}\pi {r^3})}}{{4\pi }}} \right)^{\frac{1}{3}}}\\\\ = {\left( {\frac{{4\pi {r^3}}}{{4\pi }}} \right)^{\frac{1}{3}}} = {({r^3})^{\frac{1}{3}}} = {r^{\frac{3}{3}}} = r\end{array}\)

ب

\(\begin{array}{l}r = {\left( {\frac{{3v}}{{4\pi }}} \right)^{\frac{1}{3}}}\quad \mathop = \limits^{(v = \frac{{32}}{3}\pi )} \quad {\left( {\frac{{3(\frac{{32}}{3}\pi )}}{{4\pi }}} \right)^{\frac{1}{3}}} = {\left( {\frac{{32\pi }}{{4\pi }}} \right)^{\frac{1}{3}}}\\\\ = {(8)^{\frac{1}{3}}} = {({2^3})^{\frac{1}{3}}} = {2^{\frac{3}{3}}} = 2\end{array}\)

الف

\(D = {({12^2} + {4^2} + {3^2})^{\frac{1}{2}}} = \sqrt {144 + 16 + 9} = \sqrt {169} = 13\)

ب

\(D = {({1^2} + {1^2} + {1^2})^{\frac{1}{2}}} = \sqrt {1 + 1 + 1} = \sqrt 3 \)