Question

Simplify the expression. Write your answers in positive exponents only.

Answer 1

after solving $\left(\frac{\left2x^{-7}y^7\right}{5x^2y^{-6}}\right)^{-3}$ we get $\frac{125x^{27}}{8y^{39}}$

Step-by-step explanation:

We need to simplify the expression:

$\left(\frac{\left2x^{-7}y^7\right}{5x^2y^{-6}}\right)^{-3}$

Solving

$=\frac{1}{\left(\frac{\left2x^{-7}y^7\right}{5x^2y^{-6}}\right)^{3}}\\\\=\frac{1}{\left(\frac{\left2y^{7+6}\right}{5x^{2+7}}\right)^{3}}\\=\frac{1}{\left(\frac{\left2y^{13}\right}{5x^{9}}\right)^{3}}\\=\frac{1}{\left(\frac{\left2^3y^{13*3}\right}{5^3x^{9*3}}\right)}\\=\frac{1}{\left(\frac{\left8y^{39}\right}{125x^{27}}\right)}\\We\,\,know\,\,\frac{1}{\frac{b}{a}}=\frac{a}{b}\\Applying the rule:\\=\frac{125x^{27}}{8y^{39}}$

So, after solving $\left(\frac{\left2x^{-7}y^7\right}{5x^2y^{-6}}\right)^{-3}$ we get $\frac{125x^{27}}{8y^{39}}$

Keywords: Exponents in fractions

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