Question

Help me out please. I need ASAP

Answer 1

Answer:

$\displaystyle \frac{3-\sqrt{27}}{27}$

Step-by-step explanation:

Simplifying Roots

When roots are found in an algebraic expression, it's convenient to recall these properties:

$\displaystyle \sqrt[m]{x^n}=\ x^{\frac{n}{m}}$

$\displaystyle x^m.\ x^n=\ x^{m+n}$

$\displaystyle (x^m)^n=\ x^{m.n}$

The expression is given as

$\displaystyle \frac{\sqrt[4]{9}-\sqrt{9}}{\sqrt[4]{9^5}}$

We know that $9=3^2$, so

$\displaystyle \frac{\sqrt[4]{3^2}-\sqrt{3^2}}{\sqrt[4]{3^{10}}}$

Applying the root property

$\displaystyle \frac{3^\frac{2}{4}-3^{\frac{2}{2}}}{3^\frac{10}{4}}$

Simplifying the fractions

$\displaystyle \frac{3^\frac{1}{2}-3^1}{3^\frac{5}{2}}$

Multiplying both parts by $3^{1/2}$

$\displaystyle \frac{3^\frac{1}{2}(3^\frac{1}{2}-3^1)}{3^\frac{1}{2}\ 3^\frac{5}{2}}$

Operating the exponents

$\displaystyle \frac{3^{\frac{1}{2}+\frac{1}{2}}-3^{1+\frac{1}{2}}}{3^{\frac{1}{2}+\frac{5}{2}}}$

Or equivalently

$\displaystyle \frac{3^1-3^\frac{3}{2}}{3^\frac{6}{2}}$

Simplifying and converting back to root notation

$\displaystyle \frac{3-\sqrt{3^3}}{3^3}$

Operating

$\boxed{\displaystyle \frac{3-\sqrt{27}}{27}}$