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(5^ncdot 10^2n-1cdot 2^2n+1)/(20^3n)

Pergunta

(5^ncdot 10^2n-1cdot 2^2n+1)/(20^3n)

(5^ncdot 10^2n-1cdot 2^2n+1)/(20^3n)

Solução

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JoanaVeterano · Tutor por 9 anos

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To simplify the expression $\frac {5^{n}\cdot 10^{2n-1}\cdot 2^{2n+1}}{20^{3n}}$, we can start by expressing everything in terms of base 2 and base 5.<br /><br />First, let's rewrite the expression using the fact that $10 = 2 \cdot 5$ and $20 = 2^2 \cdot 5$:<br /><br />$\frac {5^{n}\cdot (2 \cdot 5)^{2n-1}\cdot 2^{2n+1}}{(2^2 \cdot 5)^{3n}}$<br /><br />Now, let's simplify the expression by distributing the exponents:<br /><br />$\frac {5^{n}\cdot 2^{2n-1}\cdot 5^{2n-1}\cdot 2^{2n+1}}{2^{6n}\cdot 5^{3n}}$<br /><br />Next, let's combine the terms with the same base:<br /><br />$\frac {5^{n+2n-1}\cdot 2^{2n-1+2n+1}}{2^{6n}\cdot 5^{3n}}$<br /><br />Simplifying the exponents, we get:<br /><br />$\frac {5^{3n-1}\cdot 2^{4n}}{2^{6n}\cdot 5^{3n}}$<br /><br />Now, let's simplify the fraction by canceling out the common terms:<br /><br />$\frac {5^{3n-1}}{5^{3n}} \cdot \frac {2^{4n}}{2^{6n}}$<br /><br />Using the property of exponents that $\frac{a^m}{a^n} = a^{m-n}$, we can simplify further:<br /><br />$5^{3n-1-3n} \cdot 2^{4n-6n}$<br /><br />Simplifying the exponents, we get:<br /><br />$5^{-1} \cdot 2^{-2n}$<br /><br />Finally, we can rewrite the expression using positive exponents:<br /><br />$\frac{1}{5} \cdot \frac{1}{2^{2n}}$<br /><br />Multiplying the fractions, we get the final simplified expression:<br /><br />$\frac{1}{5 \cdot 2^{2n}}$<br /><br />Therefore, the simplified form of the given expression is $\frac{1}{5 \cdot 2^{2n}}$.
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