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Степеновање – решени задаци 3

Задаци

Текст задатака објашњених у видео лекцији.

Пр.5   Упрости дати израз: $\frac{{{{\left( {a{b^{ - 1}} + 1} \right)}^2}}}{{a{b^{ - 1}} - {a^{ - 1}}b}}.\frac{{{a^3}{b^{ - 3}} - 1}}{{{a^2}{b^{ - 2}} + a{b^{ - 1}} + 1}}:\frac{{{a^3}{b^{ - 3}} + 1}}{{a{b^{ - 1}} + {a^{ - 1}}b - 1}} = $.

Пр.5

\[\begin{gathered}
\frac{{{{\left( {a{b^{ - 1}} + 1} \right)}^2}}}{{a{b^{ - 1}} - {a^{ - 1}}b}}.\frac{{{a^3}{b^{ - 3}} - 1}}{{{a^2}{b^{ - 2}} + a{b^{ - 1}} + 1}}:\frac{{{a^3}{b^{ - 3}} + 1}}{{a{b^{ - 1}} + {a^{ - 1}}b - 1}} = \hfill \\
\frac{{{{\left( {a\frac{1}{b} + 1} \right)}^2}}}{{a\frac{1}{b} - \frac{1}{a}b}}.\frac{{{a^3}\frac{1}{{{b^3}}} - 1}}{{{a^2}\frac{1}{{{b^2}}} + a\frac{1}{b} + 1}}:\frac{{{a^3}\frac{1}{{{b^3}}} + 1}}{{a\frac{1}{b} + \frac{1}{a}b - 1}} = \hfill \\
= \frac{{{{\left( {\frac{a}{b} + 1} \right)}^2}}}{{\frac{a}{b} - \frac{b}{a}}}.\frac{{\frac{{{a^3}}}{{{b^3}}} - 1}}{{\frac{{{a^2}}}{{{b^2}}} + \frac{a}{b} + 1}}:\frac{{\frac{{{a^3}}}{{{b^3}}} + 1}}{{\frac{a}{b} + \frac{b}{a} - 1}} = \hfill \\
= \frac{{{{\left( {\frac{{a + b}}{b}} \right)}^2}}}{{\frac{{{a^2} - {b^2}}}{{ab}}}}.\frac{{\frac{{{a^3} - {b^3}}}{{{b^3}}}}}{{\frac{{{a^2} + ab + {b^2}}}{{{b^2}}}}}:\frac{{\frac{{{a^3} + {b^3}}}{{{b^3}}}}}{{\frac{{{a^2} - ab + {b^2}}}{{ab}}}} = \hfill \\
= \frac{{\frac{{{{\left( {a + b} \right)}^2}}}{{{b^2}}}}}{{\frac{{\left( {a - b} \right)\left( {a + b} \right)}}{{ab}}}}.\frac{{{b^2}\left( {{a^3} - {b^3}} \right)}}{{{b^3}\left( {{a^2} + ab + {b^2}} \right)}}:\frac{{ab\left( {{a^3} + {b^3}} \right)}}{{{b^3}\left( {{a^2} - ab + {b^2}} \right)}} = \hfill \\
= \frac{{ab{{\left( {a + b} \right)}^2}}}{{{b^2}\left( {a - b} \right)\left( {a + b} \right)}}.\frac{{\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)}}{{b\left( {{a^2} + ab + {b^2}} \right)}} \cdot \frac{{{b^3}\left( {{a^2} - ab + {b^2}} \right)}}{{ab\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}} = \hfill \\
= 1 \hfill \\
\end{gathered} \]

Услови: 

\[\begin{gathered}
a \ne 0 \hfill \\
b \ne 0 \hfill \\
a \ne \pm b \hfill \\
\end{gathered} \]

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