Question

I dont understand this

Answer 1

Answer:

$\dfrac{1}{2x(x-1)}$

Step-by-step explanation:

Given

$\dfrac{x^2+2x+1}{x^2-1}\div (2x^2+2x)$

Consider the numerator:

$x^2+2x+1=(x+1)^2$

Consider the denominator:

$x^2-1=(x-1)(x+1)$

Hence, the fraction becomes

$\dfrac{(x+1)^2}{(x-1)(x+1)}=\dfrac{x+1}{x-1}$

Consider the expression in brackets:

$2x^2+2x=2x(x+1)$

Divide:

$\dfrac{x^2+2x+1}{x^2-1}\div (2x^2+2x)=\dfrac{x+1}{x-1}\div 2x(x+1)=\dfrac{x+1}{x-1}\times \dfrac{1}{2x(x+1)}=\dfrac{1}{2x(x-1)}$