Răspuns:
Explicație pas cu pas:
[tex]\dfrac{x+1}{x^2+1}:\left(\dfrac{x+3}{4x-4}-\dfrac{1}{x-1}\right)=\\\\\dfrac{x+1}{x^2+1}:\left(\dfrac{x+3}{4(x-1)}-\dfrac{1}{x-1}\right)=\\\dfrac{x+1}{x^2+1}:\dfrac{x+3-4}{4(x-1)}=\\\dfrac{x+1}{x^2+1}:\dfrac{x-1}{4(x-1)}=\\\dfrac{x+1}{x^2+1}:\dfrac{1}{4}=\boxed{\dfrac{4x+4}{x^2+1}}[/tex]
