Question

sume all variables are positive.
11.) \sqrt{(\sqrt4{16x} - \sqrt4{x}

Explanation:

Step1: Rewrite roots as exponents

Recall that $\sqrt[n]{a}=a^{\frac{1}{n}}$, so:
$\sqrt[4]{16x}=16^{\frac{1}{4}}x^{\frac{1}{4}}$, $\sqrt[4]{x}=x^{\frac{1}{4}}$
The expression becomes $\sqrt{16^{\frac{1}{4}}x^{\frac{1}{4}} - x^{\frac{1}{4}}}$

Step2: Simplify $16^{\frac{1}{4}}$

Since $16=2^4$, $16^{\frac{1}{4}}=(2^4)^{\frac{1}{4}}=2$
Substitute back: $\sqrt{2x^{\frac{1}{4}} - x^{\frac{1}{4}}}$

Step3: Combine like terms

Factor out $x^{\frac{1}{4}}$ inside the square root:
$\sqrt{x^{\frac{1}{4}}(2 - 1)}=\sqrt{x^{\frac{1}{4}}}$

Step4: Rewrite square root as exponent

$\sqrt{x^{\frac{1}{4}}}=(x^{\frac{1}{4}})^{\frac{1}{2}}$

Step5: Simplify the exponent

Multiply exponents: $(x^{\frac{1}{4}})^{\frac{1}{2}}=x^{\frac{1}{4}\times\frac{1}{2}}=x^{\frac{1}{8}}$
Or rewrite as a root: $x^{\frac{1}{8}}=\sqrt[8]{x}$

Answer:

$\sqrt[8]{x}$ (or $x^{\frac{1}{8}}$)