Answer:
[tex]\large\boxed{\dfrac{1}{27}}[/tex]
Step-by-step explanation:
[tex]\dfrac{\left(\frac{1}{3}\right)^2\cdot\left(\frac{1}{3}\right)^{-7}}{\left[\left(\frac{1}{3}\right)^4\right]^{-2}}\\\\\text{use}\\\\a^n\cdot a^m=a^{n+m}\\\\\dfrac{a^n}{a^m}=a^{n-m}\\\\(a^n)^m=a^{nm}\\\\=\dfrac{\left(\frac{1}{3}\right)^{2+(-7)}}{\left(\frac{1}{3}\right)^{(4)(-2)}}=\dfrac{\left(\frac{1}{3}\right)^{-5}}{\left(\frac{1}{3}\right)^{-8}}=\left(\dfrac{1}{3}\right)^{-5-(-8)}\\\\=\left(\dfrac{1}{3}\right)^{-5+8}=\left(\dfrac{1}{3}\right)^3=\dfrac{1^3}{3^3}=\dfrac{1}{27}[/tex]
