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.