Ta có \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\dfrac{\sqrt{9\left(x+y\right)^2}}{\sqrt{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}.\dfrac{3\left|x+y\right|}{2}=\dfrac{3\left|x+y\right|}{\left(x-y\right)\left(x+y\right)}\)Nếu x+y>0 thì
\(\dfrac{3\left|x+y\right|}{\left(x-y\right)\left(x+y\right)}=\dfrac{3}{x-y}\)
Nếu x+y<0 thì
\(\dfrac{3\left|x+y\right|}{\left(x-y\right)\left(x+y\right)}=\dfrac{-3}{x-y}\)