Answer
$2x^2\sqrt{5}$
Work Step by Step
RECALL:
(i) For any non-negative real numbers a and b,
$\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ and $\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$.
(ii) For any non-negative real number a, $\sqrt{a^2}=a$.
(iii) $\dfrac{a^m}{a^n} = a^{m-n}, a\ne0$.
(iv) $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$.
Use rule (i) above to obtain
$\require{cancel}
\sqrt{\dfrac{200x^3}{10x^{-1}}}
\\=\sqrt{\dfrac{\cancel{200}20x^3}{\cancel{10}x^{-1}}}.$
Use rule (iii) above to obtain
$\sqrt{20x^{3-(-1)}}
\\=\sqrt{20x^{3+1}}
\\=\sqrt{20x^4}.$
Factor the radicand so that one factor is a perfect square to obtain
$\sqrt{4x^4(5)}
\\=\sqrt{(2x^2)^2(5)}.$
Use rule (iv) above to obtain
$\\\sqrt{(2x^2)^2} \cdot \sqrt{5}.$
Simplify to obtain
$2x^2\sqrt{5}.$
