Respuesta :

Answer:

D

Step-by-step explanation:

Using the rule of radicals

[tex]\frac{\sqrt{a} }{\sqrt{b} }[/tex] ⇔ [tex]\sqrt{\frac{a}{b} }[/tex]

given

[tex]\frac{\sqrt{30(x-1)} }{\sqrt{5(x-1)^2} }[/tex]

= [tex]\sqrt{\frac{30(x-1)}{5(x-1)^2} }[/tex]

[ cancel 30 and 5 by 5 and (x - 1) / (x - 1)² by (x - 1) ]

= [tex]\sqrt{\frac{6}{x-1} }[/tex] → D

Answer:

D. [tex]\sqrt\frac{6}{x-1}[/tex]

Step-by-step explanation:

Given:

[tex]\frac{\sqrt{30(x-1)} }{\sqrt{5(x-1)^2} }[/tex]

We can write [tex]\sqrt{xy} = \sqrt{x} \sqrt{y}[/tex]

Using this property we can rewrite the given expression as

= [tex]\frac{\sqrt{30}\sqrt{x-1}  }{\sqrt{5} \sqrt{x-1}\sqrt{x-1}  }[/tex]

Now we can simplify √30/√5 = √6 and we can cancel out √(x - 1) both in the numerator and in the denominator, so we get

= [tex]\sqrt\frac{6}{x-1}[/tex]

Therefore, the answer is D. [tex]\sqrt\frac{6}{x-1}[/tex]