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Кореновање - решени задаци 1

Примена правила за кореновање на једноставним примерима.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.1   Упростити израз:

          ${\left( {\frac{{16}}{{25}}} \right)^{\frac{1}{2}}} =$

Пр.2   ${\left( {2\frac{1}{4}} \right)^{ - \frac{1}{2}}} = $

Пр.3   ${\left( {\frac{9}{{16}}} \right)^{ - 10}}:{\left( {\frac{4}{3}} \right)^{ - \frac{4}{5}}} = $

Пр.4   ${\left( {{{\left( {1\frac{9}{{16}}} \right)}^{ - \frac{1}{2}}} - {{\left( {1 - \frac{{16}}{{25}}} \right)}^{ - \frac{1}{2}}}} \right)^{ - 1}} = $

Пр.5   $\sqrt[3]{{81 \cdot {x^5}{y^{10}}}} = $

Пр.1 \[{\left( {\frac{{16}}{{25}}} \right)^{\frac{1}{2}}} = \sqrt {\frac{{16}}{{25}}}  = \frac{{\sqrt {16} }}{{\sqrt {25} }} = \frac{4}{5}\]

Пр.2 \[{\left( {2\frac{1}{4}} \right)^{ - \frac{1}{2}}} = {\left( {\frac{9}{4}} \right)^{ - \frac{1}{2}}} = {\left( {\frac{4}{9}} \right)^{\frac{1}{2}}} = \sqrt {\frac{4}{9}}  = \frac{{\sqrt 4 }}{{\sqrt 9 }} = \frac{2}{3}\]

Пр.3 

\[\begin{gathered}
{\left( {\frac{9}{{16}}} \right)^{ - 10}}:{\left( {\frac{4}{3}} \right)^{ - \frac{4}{5}}} = {\left( {\frac{{16}}{9}} \right)^{10}}:{\left( {\frac{3}{4}} \right)^{\frac{4}{5}}} = \frac{{{{16}^{10}}}}{{{9^{10}}}}:\frac{{{3^{\frac{4}{5}}}}}{{{4^{\frac{4}{5}}}}} = \frac{{{{\left( {{4^2}} \right)}^{10}}}}{{{{\left( {{3^2}} \right)}^{10}}}} \cdot \frac{{{4^{\frac{4}{5}}}}}{{{3^{\frac{4}{5}}}}} = \hfill \\
= \frac{{{4^{20}}}}{{{3^{20}}}} \cdot \frac{{{4^{\frac{4}{5}}}}}{{{3^{\frac{4}{5}}}}} = \frac{{{4^{20}} \cdot {4^{\frac{4}{5}}}}}{{{3^{20}} \cdot {3^{\frac{4}{5}}}}} = \frac{{{4^{\frac{{104}}{5}}}}}{{{3^{\frac{{104}}{5}}}}} = {\left( {\frac{4}{3}} \right)^{\frac{{104}}{5}}} = \sqrt[5]{{{{\left( {\frac{4}{3}} \right)}^{104}}}} \hfill \\
\end{gathered} \]

Пр.4 

\[\begin{gathered}
{\left( {{{\left( {1\frac{9}{{16}}} \right)}^{ - \frac{1}{2}}} - {{\left( {1 - \frac{{16}}{{25}}} \right)}^{ - \frac{1}{2}}}} \right)^{ - 1}} = {\left( {{{\left( {\frac{{25}}{{16}}} \right)}^{ - \frac{1}{2}}} - {{\left( {\frac{9}{{25}}} \right)}^{ - \frac{1}{2}}}} \right)^{ - 1}} = \hfill \\
= {\left( {{{\left( {\frac{{16}}{{25}}} \right)}^{\frac{1}{2}}} - {{\left( {\frac{{25}}{9}} \right)}^{\frac{1}{2}}}} \right)^{ - 1}} = {\left( {\sqrt {\left( {\frac{{16}}{{25}}} \right)} - \sqrt {\left( {\frac{{25}}{9}} \right)} } \right)^{ - 1}} = {\left( {\frac{4}{5} - \frac{5}{3}} \right)^{ - 1}} = \hfill \\
= {\left( {\frac{{12 - 25}}{{15}}} \right)^{ - 1}} = {\left( { - \frac{{13}}{{15}}} \right)^{ - 1}} = - \frac{{15}}{{13}} \hfill \\
\end{gathered} \]

Пр.5 \[\sqrt[3]{{81 \cdot {x^5}{y^{10}}}} = \sqrt[3]{{{3^4}{x^6}{y^{10}}}} = \sqrt[3]{{3 \cdot {3^3} \cdot {x^3} \cdot {x^3} \cdot {y^9} \cdot y}} = 3{x^2}{y^3}\sqrt[3]{{3y}}\]