To simplify the expression
\[ \frac{t-5}{1-t} - \frac{t+4}{t+1} + \frac{t+2}{t^2-1}, \]
we first notice that \( t^2 - 1 \) can be factored:
\[ t^2 - 1 = (t-1)(t+1). \]
This means the common denominator for all terms in the expression is \( (1-t)(t+1)(t-1) \) or, equivalently, \( -(t-1)(t+1) \) since \( 1-t = -(t-1) \).
Next, we rewrite each term with the common denominator.
- The first term becomes:
\[ \frac{t-5}{1-t} = \frac{t-5}{-(t-1)} = -\frac{t-5}{t-1}. \] To express with the common denominator, we multiply by \(\frac{(t+1)(t-1)}{(t+1)(t-1)}\):
\[ \frac{-(t-5)(t+1)}{(t-1)(t+1)}. \]
- The second term is:
\[ -\frac{t+4}{t+1} = -\frac{(t+4)(1-t)}{(t+1)(1-t)} = -\frac{(t+4)(-t+1)}{(1-t)(t+1)}. \]
As above, we multiply by \(\frac{(t-1)(t+1)}{(t-1)(t+1)}\):
\[ = -\frac{-(t+4)(t-1)}{(t-1)(t+1)} = \frac{(t+4)(t-1)}{(t-1)(t+1)}. \]
- The third term becomes:
\[ \frac{t+2}{t^2-1} = \frac{t+2}{(t-1)(t+1)}. \] For this too we multiply by \(\frac{(1-t)}{(1-t)}\):
\[ \frac{(t+2)(1-t)}{(1-t)(t-1)(t+1)}. \]
Now we can write the entire expression as a single fraction:
\[ \frac{-(t-5)(t+1) + (t+4)(t-1) + (t+2)(1-t)}{(1-t)(t-1)(t+1)}. \]
Now we simplify the numerator:
- Expand \( -(t-5)(t+1) \):
\[ -(t^2 + t - 5t - 5) = -t^2 + 4t + 5. \]
- Expand \( (t+4)(t-1) \):
\[ t^2 - t + 4t - 4 = t^2 + 3t - 4. \]
- Expand \( (t+2)(1-t) \):
\[ t + 2 - t^2 - 2t = -t^2 - t + 2. \]
Now combine these results:
\[ (-t^2 + 4t + 5) + (t^2 + 3t - 4) + (-t^2 - t + 2). \]
Combine like terms:
\[ (-t^2 + t^2 - t^2) + (4t + 3t - t) + (5 - 4 + 2), \]
which simplifies to:
\[ 0 + 6t + 3 = 6t + 3. \]
This makes our simplified expression:
\[ \frac{6t + 3}{(1-t)(t-1)(t+1)}. \]
Next, factor out a 3 from the numerator:
\[ \frac{3(2t + 1)}{(1-t)(t-1)(t+1)}. \]
Since \( 1-t = -(t-1) \), we rewrite the denominator:
\[ \frac{3(2t + 1)}{-(t-1)^2(t+1)}. \]
Final simplified expression:
\[ \frac{-3(2t + 1)}{(t-1)^2(t+1)}. \]
Thus, the final result is
\[ \frac{-3(2t + 1)}{(t - 1)^2(t + 1)}. \]