Answer
$\frac{x-4}{12x^2-84x+72}$
Work Step by Step
All fractions in the denominator have the least common denominator (LCD) of $3(x-1)(x+1)(x-4)$.
Simplify the complex fraction:
$\frac{\frac{1}{3x^2-3}}{\frac{5}{x+1}-\frac{x+4}{x^2-3x-4}}=\frac{\frac{1}{3(x+1)(x-1)}}{\frac{5}{x+1}-\frac{x+4}{(x+1)(x-4)}}$ (Multiply the numerator and denominator by the LCD)
$=\frac{\frac{1}{3(x+1)(x-1)}}{\frac{5}{x+1}-\frac{x+4}{(x+1)(x-4)}}\times \frac{3(x-1)(x+1)(x-4)}{3(x-1)(x+1)(x-4)}$
$=\frac{x-4}{5\times 3(x-1)(x-4)-(x+4)\times 3(x-1)}$ (Simplify the denominator)
$=\frac{x-4}{15x^2-75x+60-(3x^2+9x-12)}$
$=\frac{x-4}{12x^2-84x+72}$
