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}
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
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 \( \frac{27}{125} x^{3} - \frac{27}{25} x^{2} + 9x - 125 \).
Answered by UpStudy AI and reviewed by a Professional Tutor
UpStudy ThothAI
Self-Developed and Ever-Improving
Thoth AI product is constantly being upgraded and optimized.
Covers All Major Subjects
Capable of handling homework in math, chemistry, biology, physics, and more.
Instant and Accurate
Provides immediate and precise solutions and guidance.
Try Now
Ask Tutors
Ask AI
10x
Fastest way to Get Answers & Solutions
By text
Enter your question here…
By image
Re-Upload
Submit