решите номер 3 и номер 4 пожалуйста))))))))))))). Created by 1margarita4. algebra-ru

решите номер 3 и номер 4 пожалуйста)))))))))))))...

August 2021 1 16 Report

решите номер 3 и номер 4 пожалуйста)))))))))))))

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13iul №3.
а)
$2 * 4^{-2} + (\frac{2}{3})^{-3} + (\frac{1}{5})^0 = 2 * \frac{1}{16} + (\frac{3}{2})^3 - 1 = \frac{1}{8} + \frac{27}{8} - 1 = 20/8 = 2,5$

б)
$\frac{3}{4} * 25^{\frac{1}{2}} + 64 ^{-\frac{1}{3}} = \frac{3}{4} * \sqrt{25} + \sqrt[3]{\frac{1}{64}} = \frac{3}{4} * 5 + \frac{1}{4} = \frac{15}{4} + \frac{1}{4} = \frac{16}{4} = 4$

в)
$12 * 3^{-3} + (\frac{3}{5})^{-2} + (\frac{1}{2}) ^0 = 12 * (\frac{1}{3})^3 + (\frac{5}{3})^2 + 1 = \frac{12+75+27}{27} = \frac{114}{27} = 4\frac{2}{9}$

г)
$\frac{2}{5}* 8^\frac{2}{3} - 25^{-\frac{1}{2}} = \frac{2}{5} * \sqrt[3]{8^2} - \sqrt{\frac{1}{25}} = \frac{2}{5}*4 - \frac{1}{5} = \frac{7}{5} = 1,4$

д)
$(\frac{3}{5})^{-8} * (\frac{5}{3})^{-6} - 3^{-11}*(\frac{1}{3})^{-9} = (\frac{5}{3})^{8} * (\frac{5}{3})^{-6} - (\frac{1}{3})^{11}*(\frac{1}{3})^{-9} = \\
\\
=(\frac{5}{3})^{2} - (\frac{1}{3})^{2} = \frac{25}{9} - \frac{1}{9} = \frac{24}{9} = \frac{8}{3} = 2 \frac{2}{3}$

е)
$(2\frac{1}{4})^{\frac{1}{2}} + 32^{-0,2} =\sqrt{\frac{9}{4}}+(\frac{1}{32} )^{\frac{2}{10}}= \frac{3}{2} + \sqrt[10]{ (\frac{1}{32})^2 } = \frac{3}{2} + \sqrt[10]{ \frac{1}{1024} } = \\
\\
= \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$

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