Question

Which expression is equivalent to Assume x 0 and y > 0.

Answer 1

The answer is the last option. 

Answer 2

Given radical expression: $\sqrt{\frac{25x^9y^3}{64x^6y^{11}}}$

$\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}\:=\:x^{a-b}$

$\frac{x^9}{x^6}=x^{9-6}=x^3$

$\frac{y^3}{y^{11}}=\frac{1}{y^{11-3}}=\frac{1}{y^8}$

$\sqrt{\frac{25x^9y^3}{64x^6y^{11}}} =\sqrt{\frac{25x^3}{64y^8}}$

$=\frac{\sqrt{25}\sqrt{x^3}}{\sqrt{64}\sqrt{y^8}}$

$\sqrt{25}=5$

$\sqrt{64}=8$

$=\frac{5\sqrt{x^3}}{8\sqrt{y^8}}$

$\sqrt{x^3}=x\sqrt{x}$

$\sqrt{y^8}=y^4$

Therefore,

$\frac{5\sqrt{x^3}}{8\sqrt{y^8}} =\frac{5x\sqrt{x}}{8y^4}$.

Therefore, $\sqrt{\frac{25x^9y^3}{64x^6y^{11}}}=\frac{5x\sqrt{x}}{8y^4}.$