Question

(t-5)/(1-t)-(t+4)/(t+1)+(t+2)/(t^(2)-1). Simplify

Answer 1

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.

1. 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 \(\frac{(t+1)(t-1)}{(t+1)(t-1)}\):

\[
\frac{-(t-5)(t+1)}{(t-1)(t+1)}.
\]

2. 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 \(\frac{(t-1)(t+1)}{(t-1)(t+1)}\):

\[
= -\frac{-(t+4)(t-1)}{(t-1)(t+1)} = \frac{(t+4)(t-1)}{(t-1)(t+1)}.
\]

3. 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:

1. Expand \( -(t-5)(t+1) \):

\[
-(t^2 + t - 5t - 5) = -t^2 + 4t + 5.
\]

2. Expand \( (t+4)(t-1) \):

\[
t^2 - t + 4t - 4 = t^2 + 3t - 4.
\]

3. 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

\[
\frac{-3(2t + 1)}{(t - 1)^2(t + 1)}.
\]