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[tex]\mathsf{\dfrac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y}}}[/tex]
Multiply and divide by the conjugate of the denominator, which is [tex]\mathsf{\sqrt{a}+2\sqrt{y}:}[/tex]
[tex]\mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}\\\\\\ \mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)^2}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}[/tex]
The numerator is a square of a sum, and the denominator is the product of a difference and a sum (product of two conjugates). Then, expand the common products:
[tex]\mathsf{=\dfrac{\big(\sqrt{a}\big)^2+2\cdot \sqrt{a}\cdot 2\sqrt{y}+\big(2\sqrt{y}\big)^2}{\big(\sqrt{a}\big)^2-\big(2\sqrt{y}\big)^2}}\\\\\\ \mathsf{=\dfrac{a+4\cdot \sqrt{a}\cdot \sqrt{y}+2^2\cdot \big(\sqrt{y}\big)^2}{a-2^2\cdot \big(\sqrt{y}\big)^2}}[/tex]
[tex]\mathsf{=\dfrac{a+4\sqrt{ay}+4y}{a-4y}}\quad\longleftarrow\quad\textsf{this is the answer.}[/tex]
I hope this helps. =)
