Problems from Operations with algebraic fractions

Development:

Firstly, we will compute the difference. To do so, we have to obtain common denominator:

$lcm\{x^2-4,x-2\}=lcm\{(x-2)\cdot (x+2),x-2\}=(x-2)\cdot (x+2)$

${\displaystyle \frac{(x-2)\cdot (x+2)}{(x-2)\cdot (x+2)}}=1\Rightarrow 1\cdot (x-1)=x-1\Rightarrow {\displaystyle \frac{x-1}{(x-2)(x+2)}}$

${\displaystyle \frac{(x-2)\cdot (x+2)}{x-2}}=x+2\Rightarrow (x+2)\cdot 1=x+2\Rightarrow {\displaystyle \frac{x+2}{(x-2)(x+2)}}$

Now we can already compute:

${\displaystyle \frac{x-1}{x^2-4}}-{\displaystyle \frac{1}{x-2}}={\displaystyle \frac{x-1}{(x-2)(x+2)}}-{\displaystyle \frac{x+2}{(x-2)(x+2)}}={\displaystyle \frac{x-1-(x+2)}{(x-2)(x+2)}}=$

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

With this result, we complete the operation:

${\displaystyle \frac{-3}{(x-2)(x+2)}}\cdot {\displaystyle \frac{x+1}{x-1}}={\displaystyle \frac{-3\cdot (x+1)}{(x-2)(x+2)(x-1)}}={\displaystyle \frac{-3x-3}{(x^2-4)(x-1)}}=$

$={\displaystyle \frac{-3x-3}{x^2\cdot (x-1)-4\cdot (x-1)}}={\displaystyle \frac{-3x-3}{x^3-x^2-4x+4}}$

Solution:

Therefore, the result is ${\displaystyle \frac{-3x-3}{x^3-x^2-4x+4}}$

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