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Степеновање - решени задаци 1

Примена правила за степеновање на једноставнијим примерима.

Задаци

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  \[\frac{{{4^2} \cdot 81 \cdot {2^3}}}{{{{27}^2} \cdot {2^5} \cdot 4}} = \frac{{{{\left( {{2^2}} \right)}^2} \cdot {3^4} \cdot {2^3}}}{{{2^3}{{\left( {{x^5}} \right)}^3}{{\left( {{y^2}} \right)}^3}}} = \frac{{{2^4} \cdot {3^4} \cdot {2^3}}}{{{3^6} \cdot {2^5} \cdot {2^2}}} = \frac{{{2^7} \cdot {3^4}}}{{{2^7} \cdot {3^6}}} = \]

\[= \frac{{3 \cdot 3 \cdot 3 \cdot 3}}{{3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3}} = \frac{1}{{3 \cdot 3}} = \frac{1}{9}\]