Answer:
The correct option is A.
Step-by-step explanation:
The given expression is
[tex]\frac{x+2}{4x^2+5x+1}\times \frac{4x+1}{x^2-4}[/tex]
[tex]\frac{x+2}{4x^2+4x+x+1}\times \frac{4x+1}{x^2-2^2}[/tex]
[tex]\frac{x+2}{4x(x+1)+1(x+1)}\times \frac{4x+1}{(x-2)(x+2)}[/tex]
[tex]\frac{x+2}{(x+1)(4x+1)}\times \frac{4x+1}{(x-2)(x+2)}[/tex]
[tex]\frac{(x+2)(4x+1)}{(x+1)(4x+1)(x-2)(x+2)}[/tex]
Cancel out common factors.
[tex]\frac{1}{(x+1)(x-2)}[/tex]
Therefore option A is correct.
Answer:
A. [tex]\frac{1}{(r+1)\cdot (r-2)}[/tex]
Step-by-step explanation:
We have been given an expression [tex]\frac{r+2}{4r^2+5r+1}\cdot \frac{4r+1}{r^2-4}[/tex]. We are asked to simplify our given expression.
We will factor the denominators of both fractions as shown below:
[tex]4r^2+5r+1[/tex]
[tex]4r^2+4r+r+1[/tex] Splitting the middle term.
[tex](4r^2+4r)+(r+1)[/tex] Making groups.
[tex]4r(r+1)+(r+1)[/tex] Factor out 4r from 1st group.
[tex]4r(r+1)+1(r+1)[/tex] Factor out 1 from 2nd group.
[tex](4r+1)(r+1)[/tex]
Using difference of squares we will factor the 2nd denominator as:
[tex]r^2-4=r^2-2^2=(r+2)(r-2)[/tex]
Substituting these values in our given problem we will get,
[tex]\frac{r+2}{(4r+1)(r+1)}\cdot \frac{4r+1}{(r+2)(r-2)}[/tex]
After cancelling out terms we will get,
[tex]\frac{1}{(r+1)}\cdot \frac{1}{(r-2)}[/tex]
[tex]\frac{1}{(r+1)\cdot (r-2)}[/tex]
Therefore, option A is the correct choice.
