Compute the following operation with algebraic fractions: $$$\Big(\dfrac{x-1}{x^2-4}-\dfrac{1}{x-2}\Big)\cdot\dfrac{x+1}{x-1}$$$
Development:
Firstly, we will compute the difference. To do so, we have to obtain common denominator:
$$lcm\{x^2-4,x-2\}=lcm\{(x-2)\cdot(x+2),x-2\}=(x-2)\cdot(x+2)$$
$$\dfrac{(x-2)\cdot(x+2)}{(x-2)\cdot(x+2)}=1 \Rightarrow 1\cdot(x-1)=x-1 \Rightarrow \dfrac{x-1}{(x-2)(x+2)}$$
$$\dfrac{(x-2)\cdot(x+2)}{x-2}=x+2 \Rightarrow (x+2)\cdot1=x+2 \Rightarrow \dfrac{x+2}{(x-2)(x+2)}$$
Now we can already compute:
$$\dfrac{x-1}{x^2-4}-\dfrac{1}{x-2}=\dfrac{x-1}{(x-2)(x+2)}-\dfrac{x+2}{(x-2)(x+2)}=\dfrac{x-1-(x+2)}{(x-2)(x+2)}=$$
$$=\dfrac{-3}{(x-2)(x+2)}$$
With this result, we complete the operation:
$$\dfrac{-3}{(x-2)(x+2)}\cdot\dfrac{x+1}{x-1}=\dfrac{-3\cdot(x+1)}{(x-2)(x+2)(x-1)}=\dfrac{-3x-3}{(x^2-4)(x-1)}=$$
$$=\dfrac{-3x-3}{x^2\cdot(x-1)-4\cdot(x-1)}=\dfrac{-3x-3}{x^3-x^2-4x+4}$$
Solution:
Therefore, the result is $$\dfrac{-3x-3}{x^3-x^2-4x+4}$$