Can you help me to simplify this algebraic expression?

Question

How can I simplify this algebraic expression?

$$\frac{-2}{x^2-1}+\frac x{x+1} - \frac 1{x-1}$$

Answer 1

$$\begin{align*} \frac{-2}{x^2-1}+\frac{x}{x+1}-\frac{1}{x-1} &= \frac{-2}{x^2-1}+\frac{x+1}{x+1}-\frac{1}{x+1}-\frac{1}{x-1} \\ &=\frac{-2}{x^2-1}+1-\frac{x-1+x+1}{x^2-1} \\ &=1-\frac{2x+2}{x^2-1} \\ &=1-\frac{2}{x-1} \\ &=\frac{x-1+2}{x+1} \\ &=\frac{x+1}{x+1} \\ &=1 \end{align*}$$

Answer 2

You have for some $A$ and $B$ $$\frac{-2}{x^2-1}=\frac{-2}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}$$ You can then calculate $A$ and $B$ by multiplication with $x^2-1$ and comparison. After that you can simplify it.

Answer 3

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

Answer 4

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

which is as simple as it can get.To check, put $x=0$ and it simplifies to $3$ on either side of the equation.