Question

given all variables below are positive, express in simplest radical form with a rational denominator.\\(dfrac{sqrt{64x^{13}y^{4}}}{sqrt{2x^{4}y}}\\)

Explanation:

Step1: Combine the radicals

Using the property $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$, we combine the two square roots into one:
$$\sqrt{\frac{64x^{13}y^{4}}{2x^{4}y}}$$

Step2: Simplify the fraction inside the radical

Simplify the coefficients and the variables separately. For the coefficients: $\frac{64}{2} = 32$. For the $x$ terms: using the rule $\frac{x^{m}}{x^{n}}=x^{m - n}$, we have $x^{13-4}=x^{9}$. For the $y$ terms: $\frac{y^{4}}{y}=y^{4 - 1}=y^{3}$. So the expression inside the radical becomes:
$$\sqrt{32x^{9}y^{3}}$$

Step3: Simplify the square root

We can break down the factors into perfect squares and non - perfect squares.

• For the coefficient: $32 = 16\times2$, and $\sqrt{16}=4$.
• For the $x$ terms: $x^{9}=x^{8}\times x$, and $\sqrt{x^{8}}=x^{4}$ (since $(x^{4})^{2}=x^{8}$).
• For the $y$ terms: $y^{3}=y^{2}\times y$, and $\sqrt{y^{2}} = y$ (since $y>0$).

Putting it all together:
$$\sqrt{16\times2\times x^{8}\times x\times y^{2}\times y}$$
Using the property $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ ($a,b\geq0$), we get:
$$\sqrt{16}\times\sqrt{x^{8}}\times\sqrt{y^{2}}\times\sqrt{2xy}$$
Substituting the square roots of the perfect squares: $\sqrt{16} = 4$, $\sqrt{x^{8}}=x^{4}$, $\sqrt{y^{2}}=y$. So the expression simplifies to:
$$4x^{4}y\sqrt{2xy}$$

Answer:

$4x^{4}y\sqrt{2xy}$