Respuesta :

Answer:

1/a²⁷

Step-by-step explanation:

Given: [tex]$ \frac{((a^2)^{-2})^5}{a^4.a^3} $ [/tex]

We know that when the bases are same the powers can be added. i.,e.,

                                           (xᵃ)ᵇ = x ⁽ᵃ ⁺ ᵇ⁾

⇒ [tex]$ \frac{((a^2)^{-2})^5}{a^4.a^3} \implies \frac{(a^2)^{-10}}{a^7} $ [/tex]

[tex]$  \implies \frac{a^{-20}}{a^7} = a^{-20 - 7} = a^{-27} $[/tex]

Also, [tex]$ \frac{x^a}{x^b} = x^{a - b} $[/tex]

This is nothing but, [tex]$ \frac{1}{a^{27}} $[/tex].

Note that [tex]$ a $[/tex] cannot be zero here. The condition is provided in the question as well.

ACCESS MORE