9514 1404 393
Answer:
(4x +3y)/(4x -3y)
Step-by-step explanation:
[tex]\dfrac{16x^2-9y^2}{(4x -3y)^2}=\dfrac{(4x)^2-(3y)^2}{(4x-3y)^2}=\dfrac{(4x-3y)(4x+3y)}{(4x-3y)(4x-3y)}\\\\=\boxed{\dfrac{4x+3y}{4x-3y}}[/tex]
