[tex]\dfrac{4x^2-x}{16x^2-1}=\dfrac{x(4x-1)}{(4x)^2-1^2}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x(4x-1)}{(4x-1)(4x+1)}=\dfrac{x}{4x+1}\\\\Answer:\ \boxed{\dfrac{4x^2-x}{16x^2-1}=\dfrac{x}{4x+1}}[/tex]
Answer:
x/4x+1
Step-by-step explanation:
