Answer:
[tex]\dfrac{1}{2x(x-1)}[/tex]
Step-by-step explanation:
Given
[tex]\dfrac{x^2+2x+1}{x^2-1}\div (2x^2+2x)[/tex]
Consider the numerator:
[tex]x^2+2x+1=(x+1)^2[/tex]
Consider the denominator:
[tex]x^2-1=(x-1)(x+1)[/tex]
Hence, the fraction becomes
[tex]\dfrac{(x+1)^2}{(x-1)(x+1)}=\dfrac{x+1}{x-1}[/tex]
Consider the expression in brackets:
[tex]2x^2+2x=2x(x+1)[/tex]
Divide:
[tex]\dfrac{x^2+2x+1}{x^2-1}\div (2x^2+2x)=\dfrac{x+1}{x-1}\div 2x(x+1)=\dfrac{x+1}{x-1}\times \dfrac{1}{2x(x+1)}=\dfrac{1}{2x(x-1)}[/tex]
