جواب کار در کلاس صفحه 92 درس 3 ریاضی و آمار دوازدهم انسانی (الگوهای غیر خطی)

\(\begin{array}{l}{3^{\frac{1}{4}}} = \sqrt[4]{3}\\\\{7^{\frac{1}{8}}} = \sqrt[8]{7}\\\\\sqrt[3]{{25}} = {5^{\frac{2}{3}}}\\\\\sqrt[{12}]{{2/7}} = 2/{7^{\frac{1}{{12}}}}\\\\{\left( {0/31} \right)^{\frac{1}{2}}} = \sqrt[2]{{0/31}}\\\\\sqrt[{10}]{1} = {1^{\frac{1}{{10}}}} = {1^{0/1}} = 1\end{array}\)

6 ماه (نصف سال) بعد : \(100 \times {(1/2)^{\frac{1}{2}}} = 100 \times \sqrt {1/2} \)

3 سال و 6 ماه بعد : \(100 \times {(1/2)^{3\frac{1}{2}}} = 100 \times {(1/2)^{\frac{7}{2}}} = 100 \times \sqrt {1/{2^7}} \)

20 روز بعد : \(100 \times {(1/2)^{\frac{{200}}{{365}}}} = 100 \times \sqrt[{365}]{{1/{2^{200}}}}\)

1 سال و 2 ماه بعد : \(100 \times {(1/2)^{1\frac{1}{6}}} = 100 \times {(1/2)^{\frac{7}{6}}} = 100 \times \sqrt[6]{{1/{2^7}}}\)

\(\begin{array}{l}{4^{\frac{1}{2}}} = {\left( {{2^2}} \right)^{\frac{1}{2}}} = {2^{2 \times \frac{1}{2}}} = 2\\\\{125^{ - \frac{1}{3}}} = {({5^3})^{ - \frac{1}{3}}} = {5^{3 \times ( - \frac{1}{3})}} = {5^{ - 1}} = \frac{1}{5}\\\\{100^{\frac{1}{2}}} = {({10^2})^{\frac{1}{2}}} = {10^{2 \times \frac{1}{2}}} = 10\\\\{32^{\frac{1}{5}}} = {({2^5})^{\frac{1}{5}}} = {2^{5 \times \frac{1}{5}}} = 2\end{array}\)

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

\( - 4{\left( {1000} \right)^{\frac{1}{3}}} = - 4{({10^3})^{\frac{1}{3}}} = - 4 \times {10^{3 \times \frac{1}{3}}} = - 4 \times 10 = - 40\)

\({3^{\frac{1}{3}}} \times {3^{\frac{2}{3}}} = {2^{\frac{1}{3}\: + \:\frac{2}{3}}} = {2^{\frac{3}{3}}} = 2\)

\({7^{\frac{3}{4}}} \times {7^{\frac{5}{4}}} = {7^{\frac{3}{4}\: + \:\frac{5}{4}}} = {2^{\frac{8}{4}}} = {2^2} = 4\)

\(\begin{array}{l}{125^{\frac{2}{3}}} \div {125^{\frac{1}{4}}} = \;\;\:{125^{\frac{2}{3} - \frac{1}{4}}} = {125^{\frac{2}{3} - \frac{1}{4}}} = {125^{\frac{5}{{12}}}} = {({5^3})^{\frac{5}{{12}}}}\\\\ = {5^{3 \times \frac{5}{{12}}}} = {5^{\frac{5}{4}}} = 5\sqrt[4]{5}\end{array}\)

\({8^{\frac{2}{7}}} \times {\left( {1/5} \right)^{\frac{2}{7}}} = {(8 \times 1/5)^{\frac{2}{7}}} = {12^{\frac{2}{7}}} = \sqrt[7]{{{{12}^2}}} = \sqrt[7]{{144}}\)

\(\begin{array}{l}\sqrt[3]{{ - 8}} = \sqrt[3]{{{{( - 2)}^3}}} = - 2\\\\ - 2 = {( - 8)^{\frac{1}{3}}} = {( - 8)^{\frac{2}{6}}}\\\\ = \sqrt[6]{{{{( - 8)}^2}}} = \sqrt[6]{{64}} = 2\;\;\: \otimes \end{array}\)