How do you simplify $\dfrac{\sqrt{5{{k}^{4}}}+3\sqrt{2k}}{\sqrt{3{{k}^{3}}}}$ ?

Hint: We know that $\dfrac{a+b}{c}$ is equal to $\dfrac{a}{c}+\dfrac{b}{c}$ , c is a non- zero number , we can use it to write $\dfrac{\sqrt{5{{k}^{4}}}+3\sqrt{2k}}{\sqrt{3{{k}^{3}}}}$ as $\dfrac{\sqrt{5{{k}^{4}}}}{\sqrt{3{{k}^{3}}}}+\dfrac{3\sqrt{2k}}{\sqrt{3{{k}^{3}}}}$ . we know that $\dfrac{\sqrt{a}}{\sqrt{b}}$ is equal to $\sqrt{\dfrac{a}{b}}$ so we can write $\dfrac{\sqrt{5{{k}^{4}}}}{\sqrt{3{{k}^{3}}}}$ as $\sqrt{\dfrac{5{{k}^{4}}}{3{{k}^{3}}}}$ and other one also. Then we can apply the exponential formula $\dfrac{{{x}^{n}}}{{{x}^{m}}}$ is equal to ${{x}^{n-m}}$, x is not 0.

Complete step by step answer:
The given equation is $\dfrac{\sqrt{5{{k}^{4}}}+3\sqrt{2k}}{\sqrt{3{{k}^{3}}}}$
We can write $\dfrac{\sqrt{5{{k}^{4}}}+3\sqrt{2k}}{\sqrt{3{{k}^{3}}}}$ = $\dfrac{\sqrt{5{{k}^{4}}}}{\sqrt{3{{k}^{3}}}}+\dfrac{3\sqrt{2k}}{\sqrt{3{{k}^{3}}}}$ .
 we know that $\dfrac{\sqrt{a}}{\sqrt{b}}$ is equal to $\sqrt{\dfrac{a}{b}}$ , where b is not 0.
$\Rightarrow $ $\dfrac{\sqrt{5{{k}^{4}}}}{\sqrt{3{{k}^{3}}}}$ = $\sqrt{\dfrac{5{{k}^{4}}}{3{{k}^{3}}}}$ and
$\Rightarrow $ $\dfrac{3\sqrt{2k}}{\sqrt{3{{k}^{3}}}}$ = $3\sqrt{\dfrac{2k}{3{{k}^{3}}}}$
 $\Rightarrow $ $\dfrac{\sqrt{5{{k}^{4}}}+3\sqrt{2k}}{\sqrt{3{{k}^{3}}}}$ = $\sqrt{\dfrac{5{{k}^{4}}}{3{{k}^{3}}}}$ + $3\sqrt{\dfrac{2k}{3{{k}^{3}}}}$
Now we can apply the formula $\dfrac{{{x}^{n}}}{{{x}^{m}}}$ is equal to ${{x}^{n-m}}$, where x is not equal to 0.
We can say $\sqrt{\dfrac{5{{k}^{4}}}{3{{k}^{3}}}}$ + $3\sqrt{\dfrac{2k}{3{{k}^{3}}}}$ is equal to $\sqrt{\dfrac{5}{3}k}$ + $\dfrac{3}{k}\sqrt{\dfrac{2}{3}}$
Further solving we get
$\Rightarrow $ $\sqrt{\dfrac{5}{3}k}$ + $\dfrac{3}{k}\sqrt{\dfrac{2}{3}}$ = $\sqrt{\dfrac{5}{3}k}+\dfrac{\sqrt{6}}{k}$
So $\sqrt{\dfrac{5}{3}k}+\dfrac{\sqrt{6}}{k}$ is the simplified form of $\dfrac{\sqrt{5{{k}^{4}}}+3\sqrt{2k}}{\sqrt{3{{k}^{3}}}}$

Note:
Square root of any number x means square root of x. so $\sqrt{3{{k}^{3}}}$ is equal to $\sqrt{3}{{\left( x \right)}^{\dfrac{3}{2}}}$ and $\sqrt{5{{x}^{4}}}$ equal to $\sqrt{5}{{x}^{2}}$ so we can write these value and can apply the formula $\dfrac{{{x}^{n}}}{{{x}^{m}}}$ is equal to ${{x}^{n-m}}$, where x is not equal to 0, we can get the answer. Whenever there is a term like $a\sqrt{b}$ , and we want to write the whole thing under square root , we have square a and then multiply it with b. so it will be $a\sqrt{b}$ = $\sqrt{{{a}^{2}}b}$ , for example $3\sqrt{2}$ = $\sqrt{18}$ . In the above question, we can solve it for k is equal to 0, in that case the value will be in the form of $\dfrac{0}{0}$ which is not valid.