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√a +2√y
√a-2√y

Do you know how to rationalize the denominator and simply?

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$$\frac{a+4\sqrt{ay}+y }{a-4y }$$

Step-by-step explanation:

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

We know that:

-  $$(a - b)*(a + b) = a^{2} -b^{2}$$
- we can always multiply by 1
- $$\frac{\sqrt{a}+2\sqrt{y} }{\sqrt{a} +2\sqrt{y} }=1$$
- $$(a+b)^2=a^2+2ab+b^2$$

Therefore,

$$\frac{\sqrt{a}+2\sqrt{y} }{\sqrt{a}-2\sqrt{y} } *1 =\frac{\sqrt{a}+2\sqrt{y} }{\sqrt{a}-2\sqrt{y} } *\frac{\sqrt{a} +2\sqrt{y} }{\sqrt{a}+2\sqrt{y} } =\frac{(\sqrt{a}+2\sqrt{y})^2 }{(\sqrt{a})^2-(2\sqrt{y})^2 } =\frac{a+4\sqrt{ay}+y }{a-4y }$$
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$$\frac{a+4\sqrt{ay}+4y }{a-4y}$$

given:

$$\frac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y} }$$

solve for:

Rationalized denominator

Step-by-step explanation:

1. Rationalize the denominator

$$\frac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y} } * \frac{2\sqrt{y} }{2\sqrt{y} }$$

2. Simplify

$$\frac{2\sqrt{y}(\sqrt{a}+2\sqrt{y} ) }{2\sqrt{y}(\sqrt{a}-2\sqrt{y}) }$$

$$\frac{a+4\sqrt{ay}+4y }{a-4y}$$
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