Question

question
given all variables below are positive, express in simplest radical form with a rational denominator.
\\(\frac{\sqrt{10x^4y^9}}{\sqrt{2x^4y^2}}\\)

Explanation:

Step1: Combine the radicals

Using the property $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$, we can combine the two square roots into one:
$\frac{\sqrt{10x^{4}y^{9}}}{\sqrt{2x^{4}y^{2}}}=\sqrt{\frac{10x^{4}y^{9}}{2x^{4}y^{2}}}$

Step2: Simplify the fraction inside the radical

Simplify the coefficients and the variables separately. For the coefficients: $\frac{10}{2} = 5$. For the $x$ terms: $\frac{x^{4}}{x^{4}}=x^{4 - 4}=x^{0} = 1$ (since $x
eq0$ as it's positive). For the $y$ terms: $\frac{y^{9}}{y^{2}}=y^{9 - 2}=y^{7}$. So we have $\sqrt{5\times1\times y^{7}}=\sqrt{5y^{7}}$

Step3: Simplify the radical $\sqrt{5y^{7}}$

We can write $y^{7}$ as $y^{6}\times y$, where $y^{6}=(y^{3})^{2}$. So $\sqrt{5y^{7}}=\sqrt{5\times y^{6}\times y}=\sqrt{y^{6}}\times\sqrt{5y}=y^{3}\sqrt{5y}$ (since $y$ is positive, $\sqrt{y^{6}} = y^{3}$)

Answer:

$y^{3}\sqrt{5y}$