Answer
$\frac{x^2-x-8}{x^2+x-2}$
Work Step by Step
Step 1. Perform operations to the numerator:
$\frac{x+4}{x}-\frac{3}{x-2}=\frac{x^2-2x+4x-8-3x}{x(x-2)}=\frac{x^2-x-8}{x(x-2)}$
Step 2. Perform operations to the denominator:
$\frac{x}{x-2}+\frac{1}{x}=\frac{x^2+x-2}{x(x-2)}$
Step 3. Use the above results in the original expression, we have:
$\frac{\frac{x+4}{x}-\frac{3}{x-2}}{\frac{x}{x-2}+\frac{1}{x}}=\frac{\frac{x^2-x-8}{x(x-2)}}{\frac{x^2+x-2}{x(x-2)}}=\frac{x^2-x-8}{x(x-2)}\cdot\frac{x(x-2)}{x^2+x-2}=\frac{x^2-x-8}{x^2+x-2}$
