Cook Blake
06/15/2023 · Elementary School

\begin{tabular}{l|l} esión utilizando el cubo de un & \\ 112. \( \left(\frac{3}{5} x-5\right)^{3} \) \\ 113. \( \left(x^{n}-y^{m}\right)^{3} \) & \\ 114. \( \left(w^{n}+1+1\right)^{3} \) & \\ 115. \( \left(\frac{1}{2} z^{n}-2\right)^{3} \) \\ 116. \( \left(\frac{2}{27} x^{n+1}-6 x\right)^{3} \) & \\ 117. \( \left(\frac{1}{6} b+\frac{3}{45}\right)^{3} \) & S 134. Hall. \\ 118. \( \left(\frac{4}{5} m x^{2}-\frac{1}{4}\right)^{3} \) & 9.26\end{tabular}

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Para expandir la expresión \( \left(\frac{3}{5} x-5\right)^{3} \), primero necesitamos recordar la fórmula para expandir un binomio al cubo: \[ (a-b)^{3} = a^{3} - 3a^{2}b + 3ab^{2} - b^{3} \] En este caso, \( a = \frac{3}{5}x \) y \( b = 5 \). Sustituyendo estos valores en la fórmula, obtenemos: \[ \left(\frac{3}{5} x-5\right)^{3} = \left(\frac{3}{5} x\right)^{3} - 3\left(\frac{3}{5} x\right)^{2} \cdot 5 + 3\left(\frac{3}{5} x\right) \cdot 5^{2} - 5^{3} \] Ahora, simplifiquemos cada término: \[ \left(\frac{3}{5} x\right)^{3} = \frac{27}{125} x^{3} \] \[ 3\left(\frac{3}{5} x\right)^{2} \cdot 5 = \frac{27}{25} x^{2} \] \[ 3\left(\frac{3}{5} x\right) \cdot 5^{2} = 9x \] \[ 5^{3} = 125 \] Por lo tanto, la expansión de \( \left(\frac{3}{5} x-5\right)^{3} \) es: \[ \frac{27}{125} x^{3} - \frac{27}{25} x^{2} + 9x - 125 \] Por lo tanto, la respuesta es: \[ \frac{27}{125} x^{3} - \frac{27}{25} x^{2} + 9x - 125 \]

Quick Answer

La expansión de \( \left(\frac{3}{5} x-5\right)^{3} \) es \( \frac{27}{125} x^{3} - \frac{27}{25} x^{2} + 9x - 125 \).
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