Question

R(x)=f(x)−g(x), where f(x)=x+1/x+3 and g(x)=x+17/x^2−x−12.

Answer 1

Answer:

$\frac{x^2-4x-21}{(x+3)(x-4)}$

Step-by-step explanation:

R(x)=($\frac{x+1}{x+3}$)-($\frac{x+17}{x^(2)-x-12}$)

R(x)=($\frac{x+1}{x+3}$)-($\frac{x+17}{(x+3)(x-4)}$)

R(x)=$\frac{(x+1)(x-4)}{(x+3)(x-4)}$ - $\frac{x+17}{(x+3)(x-4)}$

R(x)= $\frac{x^2-3x-4-x-17}{(x+3)(x-4)}$

R(x)=$\frac{x^2-4x-21}{(x+3)(x-4)}$