Answer:
[tex]\displaystyle \frac{3-\sqrt{27}}{27}[/tex]
Step-by-step explanation:
Simplifying Roots
When roots are found in an algebraic expression, it's convenient to recall these properties:
[tex]\displaystyle \sqrt[m]{x^n}=\ x^{\frac{n}{m}}[/tex]
[tex]\displaystyle x^m.\ x^n=\ x^{m+n}[/tex]
[tex]\displaystyle (x^m)^n=\ x^{m.n}[/tex]
The expression is given as
[tex]\displaystyle \frac{\sqrt[4]{9}-\sqrt{9}}{\sqrt[4]{9^5}}[/tex]
We know that [tex]9=3^2[/tex], so
[tex]\displaystyle \frac{\sqrt[4]{3^2}-\sqrt{3^2}}{\sqrt[4]{3^{10}}}[/tex]
Applying the root property
[tex]\displaystyle \frac{3^\frac{2}{4}-3^{\frac{2}{2}}}{3^\frac{10}{4}}[/tex]
Simplifying the fractions
[tex]\displaystyle \frac{3^\frac{1}{2}-3^1}{3^\frac{5}{2}}[/tex]
Multiplying both parts by [tex]3^{1/2}[/tex]
[tex]\displaystyle \frac{3^\frac{1}{2}(3^\frac{1}{2}-3^1)}{3^\frac{1}{2}\ 3^\frac{5}{2}}[/tex]
Operating the exponents
[tex]\displaystyle \frac{3^{\frac{1}{2}+\frac{1}{2}}-3^{1+\frac{1}{2}}}{3^{\frac{1}{2}+\frac{5}{2}}}[/tex]
Or equivalently
[tex]\displaystyle \frac{3^1-3^\frac{3}{2}}{3^\frac{6}{2}}[/tex]
Simplifying and converting back to root notation
[tex]\displaystyle \frac{3-\sqrt{3^3}}{3^3}[/tex]
Operating
[tex]\boxed{\displaystyle \frac{3-\sqrt{27}}{27}}[/tex]
