Matemática, perguntado por loadingtv, 5 meses atrás

4) Aplicando as propriedades das potências, simplifique as expressões:

Anexos:

Soluções para a tarefa

Respondido por CyberKirito
1

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\tt a)\\\begin{array}{c|l}\sf256&\sf2\\\sf128&\sf2\\\sf64&\sf2\\\sf32&\sf2\\\sf16&\sf2\\\sf8&\sf2\\\sf4&\sf2\\\sf2&\sf2\\\sf1\end{array}\longrightarrow \rm 256=2^8\\\sf \dfrac{256\cdot4^9}{8^7}=\dfrac{2^8\cdot(2^2)^9}{(2^3)^7}=\dfrac{2^8\cdot2^{18}}{2^21}\\\sf=\dfrac{2^{26}}{2^{21}}=2^{26-21}=2^5=32\blue{\checkmark}

\tt b)\\\begin{array}{c|l}\sf243&\sf3\\\sf81&\sf3\\\sf27&\sf3\\\sf9&\sf3\\\sf3&\sf3\\\sf1\end{array}\longrightarrow \rm 243=3^5\\\sf\dfrac{9^3\cdot 27^4\cdot 3^{-7}}{\dfrac{1}{3}\cdot243^2}=\dfrac{(3^2)^3\cdot (3^3)^4\cdot 3^{-7}}{3^{-1}\cdot(3^5)^2}=\dfrac{3^6\cdot3^{12}\cdot 3^{-7}}{3^{-1}\cdot 3^{10}}\\\sf=\dfrac{3^{6+12-7}}{3^{-1+10}}=\dfrac{3^{11}}{3^9}=3^{11-9}=3^2=9\blue{\checkmark}

\tt c)\\\begin{array}{c|l}\sf125&\sf5\\\sf25&\sf5\\\sf 5&\sf5\\\sf1\end{array}\longrightarrow 125=5^3\\\sf\dfrac{125^6\cdot25^{-3}}{(5^2)^{-3}\cdot25^7}=\dfrac{(5^3)^6\cdot(5^2)^{-3}}{5^{-6}\cdot(5^2)^7}=\dfrac{5^{18}\cdot\diagup\!\!\!\!\!\!5^\!\!{-6}}{\diagup\!\!\!\!\!\! 5^{-6}\cdot 5^{14}}\\\sf=5^{18-14}=5^4=625\blue{\checkmark}

\tt d)\\\sf\dfrac{12\cdot10^{-3}\cdot10^{-4}\cdot10^9}{3\cdot10^{-1}\cdot10^4}=\dfrac{4\cdot\diagup\!\!\!\!3\cdot10^{-3-5+9}}{\diagup\!\!\!\!4\cdot10^{-1+4}}\\\sf=4\cdot\dfrac{10^1}{10^3}=4\cdot10^{1-3}=4\cdot10^{-2}=4\cdot0,01=0,04\blue{\checkmark}

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