Respuesta :

rdacoder

Answer:

[tex]\frac{a+4\sqrt{ay}+4y }{a-4y}[/tex]

given:

[tex]\frac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y} }[/tex]

solve for:

Rationalized denominator

Step-by-step explanation:

1. Rationalize the denominator

[tex]\frac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y} } * \frac{2\sqrt{y} }{2\sqrt{y} }[/tex]

2. Simplify

[tex]\frac{2\sqrt{y}(\sqrt{a}+2\sqrt{y} ) }{2\sqrt{y}(\sqrt{a}-2\sqrt{y}) }[/tex]

[tex]\frac{a+4\sqrt{ay}+4y }{a-4y}[/tex]

WonderlandWanderer

Answer:

[tex]\frac{a+4\sqrt{ay}+y }{a-4y }[/tex]

Step-by-step explanation:

As far as I understand, it looks like this:   [tex]\frac{\sqrt{a}+2\sqrt{y} }{\sqrt{a}-2\sqrt{y} }[/tex]

We know that:

  1. [tex](a - b)*(a + b) = a^{2} -b^{2}[/tex]  
  2. we can always multiply by 1
  3. [tex]\frac{\sqrt{a}+2\sqrt{y} }{\sqrt{a} +2\sqrt{y} }=1[/tex]  
  4. [tex](a+b)^2=a^2+2ab+b^2[/tex]

Therefore,

[tex]\frac{\sqrt{a}+2\sqrt{y} }{\sqrt{a}-2\sqrt{y} } *1 =\frac{\sqrt{a}+2\sqrt{y} }{\sqrt{a}-2\sqrt{y} } *\frac{\sqrt{a} +2\sqrt{y} }{\sqrt{a}+2\sqrt{y} } =\frac{(\sqrt{a}+2\sqrt{y})^2 }{(\sqrt{a})^2-(2\sqrt{y})^2 } =\frac{a+4\sqrt{ay}+y }{a-4y }[/tex]