Điều kiện \(x\ne y\)
\(A=\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{4}}=\frac{2}{\left(x-y\right)\left(x+y\right)}.\frac{\sqrt{3}.\left|x+y\right|}{2}=\frac{\sqrt{3}\left|x+y\right|}{\left(x-y\right)\left(x+y\right)}\)
Nếu \(x+y>0\) thì \(A=\frac{\sqrt{3}}{x-y}\)
Nếu \(x+y< 0\) thì \(A=\frac{\sqrt{3}}{y-x}\)