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Кореновање - решени задаци 2

Примена правила за кореновање на сложенијим примерима.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.6   Упростити дати израз:

           $\left( {{x^n} \cdot {x^{\frac{1}{{n - 1}}}}} \right):{\left( {{x^{{n^2}}}} \right)^{\frac{1}{{n               -  1}}}} = $

Пр.7   $x\sqrt {x\sqrt {{x^3}\sqrt x } }  \cdot \sqrt[3]{{x\sqrt x }} = $

Пр.8   ${\left( {\sqrt[5]{{{a^2}{b^3}}}} \right)^8}:{\left( {\sqrt[5]{{{a^2}{b^3}}}} \right)^3} = $

Пр.9   $\sqrt {A + \sqrt B }  = $

\[\begin{gathered}
\sqrt {A + \sqrt B } = \sqrt {\frac{{A + \sqrt {{A^2} - B} }}{2}} + \sqrt {\frac{{A - \sqrt {{A^2} - B} }}{2}} \hfill \\
A \geqslant 0 \hfill \\
B \geqslant 0 \hfill \\
{A^2} \geqslant B \hfill \\
\end{gathered} \]

Пр.6

\[\begin{gathered}
\left( {{x^n} \cdot {x^{\frac{1}{{n - 1}}}}} \right):{\left( {{x^{{n^2}}}} \right)^{\frac{1}{{n - 1}}}} = {x^{n + \frac{1}{{n - 1}}}}:{x^{{n^2} \cdot \frac{1}{{n - 1}}}} = {x^{n + \frac{1}{{n - 1}} - }}^{{n^2} \cdot \frac{1}{{n - 1}}} = {x^{\frac{{n\left( {n - 1} \right) + 1 - {n^2}}}{{n - 1}}}} = \hfill \\
= {x^{\frac{{{n^2} - n + 1 - {n^2}}}{{n - 1}}}} = {x^{\frac{{ - \left( {n - 1} \right)}}{{n - 1}}}} = {x^{ - 1}} = \frac{1}{x} \hfill \\
x \ne 0 \hfill \\
\end{gathered} \]

Пр.7

\[\begin{gathered}
x\sqrt {x\sqrt {x\sqrt[3]{x}} } \cdot \sqrt[3]{{x\sqrt x }} = x\sqrt {x\sqrt {x{x^{\frac{1}{3}}}} } \cdot \sqrt[3]{{x{x^{\frac{1}{2}}}}} = x\sqrt {x\sqrt {{x^{\frac{4}{3}}}} } \cdot \sqrt[3]{{{x^{\frac{3}{2}}}}} = \hfill \\
= x\sqrt {x{{\left( {{x^{\frac{4}{3}}}} \right)}^{\frac{1}{2}}}} \cdot {\left( {{x^{\frac{3}{2}}}} \right)^{\frac{1}{3}}} = x\sqrt {x{x^{\frac{2}{3}}}} \cdot {x^{\frac{1}{2}}} = x\sqrt {{x^{\frac{5}{3}}}} \cdot {x^{\frac{1}{2}}} = x{x^{\frac{5}{6}}} \cdot {x^{\frac{1}{2}}} = \hfill \\
= {x^{1 + \frac{5}{6} + \frac{1}{2}}} = {x^{\frac{{14}}{6}}} = {x^{\frac{7}{3}}} = \sqrt[3]{{{x^7}}} \hfill \\
\end{gathered} \]

Пр.8 \[{\left( {\sqrt[5]{{{a^2}{b^3}}}} \right)^8}:{\left( {\sqrt[5]{{{a^2}{b^3}}}} \right)^3} = {\left( {\sqrt[5]{{{a^2}{b^3}}}} \right)^5} = {a^2}{b^3}\]

Пр.9

\[{\left. {\sqrt {A + \sqrt B }  = \sqrt {\frac{{A + \sqrt {{A^2} - B} }}{2}}  + \sqrt {\frac{{A - \sqrt {{A^2} - B} }}{2}} } \right|^2}\]

\[{\left( {\sqrt {A + \sqrt B } } \right)^2} = {\left( {\sqrt {\frac{{A + \sqrt {{A^2} - B} }}{2}}  + \sqrt {\frac{{A - \sqrt {{A^2} - B} }}{2}} } \right)^2}\]

\[\begin{gathered}
A + \sqrt B = {\left( {\sqrt {\frac{{A + \sqrt {{A^2} - B} }}{2}} } \right)^2} + \hfill \\
+ 2\left( {\sqrt {\frac{{A + \sqrt {{A^2} - B} }}{2}} \sqrt {\frac{{A - \sqrt {{A^2} - B} }}{2}} } \right) + {\left( {\sqrt {\frac{{A - \sqrt {{A^2} - B} }}{2}} } \right)^2} = \hfill \\
\end{gathered} \]

\[\begin{gathered}
= \frac{{A + \sqrt {{A^2} - B} }}{2} + 2\sqrt {\frac{{\left( {A + \sqrt {{A^2} - B} } \right)\left( {A - \sqrt {{A^2} - B} } \right)}}{4}} + \hfill \\
+ \frac{{A - \sqrt {{A^2} - B} }}{2} = \hfill \\
\end{gathered} \]

\[ = \frac{{A + \sqrt {{A^2} - B}  + A - \sqrt {{A^2} - B} }}{2} + 2\sqrt {\frac{{{A^2} - {{\left( {\sqrt {{A^2} - B} } \right)}^2}}}{4}}  = \]

\[ = \frac{{2A}}{A} + 2\sqrt {\frac{{{A^2} - {A^2} - B}}{4}}  = A + 2\frac{{\sqrt B }}{2} = A + \sqrt B \]