Question

What is the difference (x/x^2+3x+2)-(1/(x+2)(x+1)

Answer 1

Answer is D i believe

Answer 2

Answer:

Option D is correct

The difference of  $\frac{x}{x^2+3x+2}- \frac{1}{(x+2)(x+1)}$ = $\frac{x-1}{x^2+3x+2}$

Step-by-step explanation:

To find the difference of the equation:

$\frac{x}{x^2+3x+2}- \frac{1}{(x+2)(x+1)}$             ....[1]

first multiply the terms

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

Substitute this in equation [1], we have

$\frac{x}{x^2+3x+2}- \frac{1}{x^2+3x+2}$

If the denominators are same, then we have

$\frac{x+(-1)}{x^2+3x+2}$ = $\frac{x-1}{x^2+3x+2}$

Therefore, the difference of the given equation  $\frac{x}{x^2+3x+2}- \frac{1}{(x+2)(x+1)}$ is,   $\frac{x-1}{x^2+3x+2}$