The product below is equivalent when x > 0 is 1/9
Resolution - Explanation
Square root is a real number x multiplied by itself - which results in a perfect value, where it is possible to calculate the real proof (which is the square root).
Given the expression, [tex]\large \sf \sqrt{\dfrac{1}{x^{2} } } \cdot \sqrt{\dfrac{x^{2} }{81} }[/tex], first step: we will calculate the root of the numerator and denominator of this fraction:
[tex]\\\large \sf \sqrt{\dfrac{1}{x^{2} } } \cdot \sqrt{\dfrac{x^{2} }{81} }[/tex]
[tex]\large \sf \dfrac{\sqrt{1} }{\sqrt{x^{2} } } \rightarrow \dfrac{1}{x}[/tex]
[tex]\large \sf \dfrac{\sqrt{x^{2} } }{\sqrt{81 } } \rightarrow \dfrac{x}{9}\\\\[/tex]
Step two: rearranging the expression and canceling the common factors x, we will have,:
[tex]\\\large \sf \dfrac{1}{\not x} \cdot \dfrac{\not x}{9}[/tex]
[tex]\pink{\boxed{\large \sf \dfrac{1}{9} }}\\[/tex]
Therefore, the final answer to this multiplication will be 1/9.
