Simplifying some square numbers expressions

Question

I need help to fix theese out. Thank you.

$ \frac { \frac {1} {\sqrt {3} } - \sqrt {12} } { \sqrt {3} } $

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$ \frac { \sqrt {x} } {\sqrt[3] {3} } + \frac {\sqrt[4] {x}} {\sqrt {x} } $

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$ \sqrt {5} - \sqrt {3} = n \Rightarrow \sqrt {5} + \sqrt {3} = ? $

Answer 1

$$ \frac { \frac {1} {\sqrt {3} } - \sqrt {12} } { \sqrt {3} } = \frac { -\frac{5}{\sqrt{3}} } { \sqrt {3} } = \frac { -5 } { \sqrt{3}\sqrt {3} } = \frac { -5 } { (\sqrt{3})^2 } = \frac { -5 } { 3 } = -\frac { 5 } { 3 } $$

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$$ \frac { \sqrt {x} } {\sqrt[3] {3} } + \frac {\sqrt[4] {x}} {\sqrt {x} } =$$ $$ \frac { \sqrt {x} } {\sqrt[3] {3} } + x^{\frac{1}{4}-\frac{1}{2}} =$$ $$ \frac { \sqrt {x} } {\sqrt[3] {3} } + x^{-\frac{1}{4}} =$$ $$ \frac { \sqrt {x} } {\sqrt[3] {3} } + \frac{1}{\sqrt[4]{x}}$$

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$$ \sqrt {5} - \sqrt {3} = n \Longleftrightarrow$$ $$\left(\sqrt{5}-\sqrt{3}\right)\cdot \left(\sqrt{5}+\sqrt{3}\right)= n\cdot \left(\sqrt{5}+\sqrt{3}\right)\Longleftrightarrow$$ $$\sqrt{5}+\sqrt{3} = \frac{2}{n}$$

Answer 2

If $$\left(\sqrt{5}-\sqrt{3}\right) = n\;,$$ Then $$\left(\sqrt{5}-\sqrt{3}\right)\cdot \left(\sqrt{5}+\sqrt{3}\right)= n\cdot \left(\sqrt{5}+\sqrt{3}\right)$$

So we get $$\displaystyle \left(\sqrt{5}+\sqrt{3}\right) = \frac{2}{n}$$