Текст задатака објашњених у видео лекцији.
Израчунати:
Пр.1 ${\left( {\frac{3}{5}} \right)^3} =$
Пр.2 ${\left( { - 2} \right)^4} =$
Пр.3 ${\left( { - 3} \right)^3} =$
Пр.4 ${\left( { - {7^4}} \right)^3} =$
Пр.5 ${\left( { - 4a{b^3}} \right)^2} = $
Пр.6 ${\left( {\frac{3}{2}} \right)^{11}} \cdot {\left( {\frac{8}{{15}}} \right)^{11}} \cdot {\left( {\frac{5}{4}} \right)^{11}} =$
Пр.7 $\frac{{{4^2} \cdot 81 \cdot {2^3}}}{{{{27}^2} \cdot {2^5} \cdot 4}} = $
Пр. 1
\[\begin{gathered}
{\left( {\frac{3}{5}} \right)^3} = \frac{{{3^3}}}{{{5^3}}} = \frac{{27}}{{125}} \hfill \\
\frac{{{3^3}}}{5} = \frac{{3 \cdot 3 \cdot 3}}{5} = \frac{{27}}{5} \hfill \\
\end{gathered} \]
Пр. 2 \[{\left( { - 2} \right)^4} = \left( { - 2} \right) \cdot \left( { - 2} \right) \cdot \left( { - 2} \right) \cdot \left( { - 2} \right) = 4 \cdot 4 = 16\]
\[ - {2^4} = - 2 \cdot 2 \cdot 2 \cdot 2 = - 16\]
Пр. 3 \[{\left( { - 3} \right)^3} = \left( { - 3} \right) \cdot \left( { - 3} \right) \cdot \left( { - 3} \right) = 9 \cdot \left( { - 3} \right) = - 27\]
\[ - {3^3} = - 27\]
Пр. 4 \[{\left( { - {7^4}} \right)^3} = - {7^{4 \cdot 3}} = - {7^{12}}\]
\[{\left( {{{\left( { - 7} \right)}^4}} \right)^3} = {\left( { - 7} \right)^{12}} = {7^{12}}\]
Пр. 5 \[{\left( { - 4a{b^3}} \right)^2} = {\left( { - 4} \right)^2}{a^2}{b^{3 \cdot 2}} = 16{a^2}{b^6}\]
\[{\left( {\frac{{3{a^2}{b^3}}}{{2{x^5}{y^2}}}} \right)^3} = \frac{{{3^3}{{\left( {{a^2}} \right)}^3}{{\left( {{b^3}} \right)}^3}}}{{{2^3}{{\left( {{x^5}} \right)}^3}{{\left( {{y^2}} \right)}^3}}} = \frac{{{3^3}{a^6}{b^9}}}{{{2^3}{x^{15}}{y^6}}}\]
Пр. 6 \[{\left( {\frac{3}{2}} \right)^{11}} \cdot {\left( {\frac{8}{{15}}} \right)^{11}} \cdot {\left( {\frac{5}{4}} \right)^{11}} = {\left( {\frac{3}{2} \cdot \frac{8}{{15}} \cdot \frac{5}{4}} \right)^{11}} = {1^{11}} = 1\]
Пр. 7 \[\frac{{{4^2} \cdot 81 \cdot {2^3}}}{{{{27}^2} \cdot {2^5} \cdot 4}} = \frac{{{{\left( {{2^2}} \right)}^2} \cdot {3^4} \cdot {2^3}}}{{{2^3}{{\left( {{x^5}} \right)}^3}{{\left( {{y^2}} \right)}^3}}} = \frac{{{2^4} \cdot {3^4} \cdot {2^3}}}{{{3^6} \cdot {2^5} \cdot {2^2}}} = \frac{{{2^7} \cdot {3^4}}}{{{2^7} \cdot {3^6}}} = \]
\[= \frac{{3 \cdot 3 \cdot 3 \cdot 3}}{{3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3}} = \frac{1}{{3 \cdot 3}} = \frac{1}{9}\]
