Pergunta
7) ((64)/(p^6))^-(1)/(3)
Solução
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Tatiana MariaProfissional · Tutor por 6 anos
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To simplify the expression (\frac{64}{p^6})^{-\frac{1}{3}}, we can follow these steps:
1. **Apply the negative exponent rule**: The negative exponent indicates taking the reciprocal of the base and then applying the positive exponent.
\left(\frac{64}{p^6}\right)^{-\frac{1}{3}} = \left(\frac{p^6}{64}\right)^{\frac{1}{3}}
2. **Apply the cube root**: Taking the cube root of both the numerator and the denominator.
\left(\frac{p^6}{64}\right)^{\frac{1}{3}} = \frac{(p^6)^{\frac{1}{3}}}{64^{\frac{1}{3}}}
3. **Simplify the numerator**: The cube root of p^6 is p^2 because (p^6)^{\frac{1}{3}} = p^{6 \cdot \frac{1}{3}} = p^2.
\frac{(p^6)^{\frac{1}{3}}}{64^{\frac{1}{3}}} = \frac{p^2}{64^{\frac{1}{3}}}
4. **Simplify the denominator**: The cube root of 64 is 4 because 64^{\frac{1}{3}} = 4.
\frac{p^2}{64^{\frac{1}{3}}} = \frac{p^2}{4}
So, the simplified form of the expression (\frac{64}{p^6})^{-\frac{1}{3}} is:
\boxed{\frac{p^2}{4}}
1. **Apply the negative exponent rule**: The negative exponent indicates taking the reciprocal of the base and then applying the positive exponent.
\left(\frac{64}{p^6}\right)^{-\frac{1}{3}} = \left(\frac{p^6}{64}\right)^{\frac{1}{3}}
2. **Apply the cube root**: Taking the cube root of both the numerator and the denominator.
\left(\frac{p^6}{64}\right)^{\frac{1}{3}} = \frac{(p^6)^{\frac{1}{3}}}{64^{\frac{1}{3}}}
3. **Simplify the numerator**: The cube root of p^6 is p^2 because (p^6)^{\frac{1}{3}} = p^{6 \cdot \frac{1}{3}} = p^2.
\frac{(p^6)^{\frac{1}{3}}}{64^{\frac{1}{3}}} = \frac{p^2}{64^{\frac{1}{3}}}
4. **Simplify the denominator**: The cube root of 64 is 4 because 64^{\frac{1}{3}} = 4.
\frac{p^2}{64^{\frac{1}{3}}} = \frac{p^2}{4}
So, the simplified form of the expression (\frac{64}{p^6})^{-\frac{1}{3}} is:
\boxed{\frac{p^2}{4}}
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