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kobusy [5.1K]
3 years ago
8

I dont understand this

Mathematics
1 answer:
Westkost [7]3 years ago
4 0

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)}

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