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Степеновање – решени задаци 2

Задаци

Пр. 1

\[{\left( { - 1\frac{1}{3}} \right)^{ - 3}} = {\left( { - \frac{4}{3}} \right)^{ - 3}} = {\left( { - \frac{3}{4}} \right)^3} =  - \frac{{{3^3}}}{{{4^3}}} =  - \frac{{27}}{{64}}\]

Пр. 2

\[{5^{ - 1}}a{b^{ - 2}} = \frac{1}{5} \cdot a \cdot \frac{1}{{{b^2}}} = \frac{a}{{5{b^2}}},b \ne 0\]

Пр. 3

1)   \[\frac{{{3^{ - 1}}{a^2}{b^{ - 3}}}}{{2{x^{ - 4}}{y^2}}} = \frac{{\frac{1}{3}{a^2}\frac{1}{{{b^3}}}}}{{2\frac{1}{{{x^4}}}{y^2}}} = \frac{{\frac{{{a^2}}}{{3{b^3}}}}}{{\frac{{2{y^2}}}{{{x^4}}}}} = \frac{{{a^2}{x^4}}}{{2{y^2} \cdot 3{b^3}}} = \frac{{{a^2}{x^4}}}{{6{y^2}{b^3}}},x,y,b \ne 0\]

2)   \[\begin{gathered}
\left( {\frac{{4{a^{ - 2}}}}{{3{b^{ - 3}}}}} \right) \cdot \frac{1}{{12{a^5}{b^2}}} = {\left( {\frac{{2bb}}{{3a}}} \right)^{ - 4}} \div {\left( {\frac{{4{b^3}}}{{3{a^2}}}} \right)^{ - 3}} \cdot \frac{1}{{12{a^5}{b^2}}} = \hfill \\
= {\left( {\frac{{3a}}{{2{b^2}}}} \right)^4} \div {\left( {\frac{{3{a^2}}}{{4{b^3}}}} \right)^3} \cdot \frac{1}{{12{a^5}{b^2}}} = \frac{{{3^4}{a^4}}}{{{2^4}{{\left( {{b^2}} \right)}^4}}} \div \frac{{{3^3}{{\left( {{a^2}} \right)}^3}}}{{{4^3}{{\left( {{b^3}} \right)}^3}}} \cdot \frac{1}{{12{a^5}{b^2}}} = \hfill \\
= \frac{{{3^4}{a^4}}}{{{2^4}{b^8}}} \cdot \frac{{{4^3}{b^9}}}{{{3^3}{a^6}}} \cdot \frac{1}{{12{a^5}{b^2}}} = \frac{{3{{\left( {{2^2}} \right)}^3}}}{{{2^4}{a^2} \cdot 12{a^5}b}} = \frac{{3 \cdot {2^6}}}{{12 \cdot {2^4}{a^7}b}} = \frac{1}{{{a^7}b}},a,b \ne 0 \hfill \\
\end{gathered} \]

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