Given radical expression: [tex]\sqrt{\frac{25x^9y^3}{64x^6y^{11}}}[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}\:=\:x^{a-b}[/tex]
[tex]\frac{x^9}{x^6}=x^{9-6}=x^3[/tex]
[tex]\frac{y^3}{y^{11}}=\frac{1}{y^{11-3}}=\frac{1}{y^8}[/tex]
[tex]\sqrt{\frac{25x^9y^3}{64x^6y^{11}}} =\sqrt{\frac{25x^3}{64y^8}}[/tex]
[tex]=\frac{\sqrt{25}\sqrt{x^3}}{\sqrt{64}\sqrt{y^8}}[/tex]
[tex]\sqrt{25}=5[/tex]
[tex]\sqrt{64}=8[/tex]
[tex]=\frac{5\sqrt{x^3}}{8\sqrt{y^8}}[/tex]
[tex]\sqrt{x^3}=x\sqrt{x}[/tex]
[tex]\sqrt{y^8}=y^4[/tex]
Therefore,
[tex]\frac{5\sqrt{x^3}}{8\sqrt{y^8}} =\frac{5x\sqrt{x}}{8y^4}[/tex].
Therefore, [tex]\sqrt{\frac{25x^9y^3}{64x^6y^{11}}}=\frac{5x\sqrt{x}}{8y^4}.[/tex]
