Answer:
Option D is correct
The difference of [tex]\frac{x}{x^2+3x+2}- \frac{1}{(x+2)(x+1)}[/tex] = [tex]\frac{x-1}{x^2+3x+2}[/tex]
Step-by-step explanation:
To find the difference of the equation:
[tex]\frac{x}{x^2+3x+2}- \frac{1}{(x+2)(x+1)}[/tex] ....[1]
first multiply the terms
[tex](x+2)(x+1) =(x^2+x+2x+2)[/tex]
Substitute this in equation [1], we have
[tex]\frac{x}{x^2+3x+2}- \frac{1}{x^2+3x+2}[/tex]
If the denominators are same, then we have
[tex]\frac{x+(-1)}{x^2+3x+2}[/tex] = [tex]\frac{x-1}{x^2+3x+2}[/tex]
Therefore, the difference of the given equation [tex]\frac{x}{x^2+3x+2}- \frac{1}{(x+2)(x+1)}[/tex] is, [tex]\frac{x-1}{x^2+3x+2}[/tex]
