<s>[INST] Simplify fraction within a fraction (Precalculus) Simplify $$\frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}}$$
My attempt:
$$=\frac{x-2}{x-2-\frac{x}{\frac{x(x-2)-(x-1)}{x-2}}} \ \ = \  \ \frac{x-2}{x-2-\frac{x}{\frac{x^2-2x-x+1}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x^2-2x}{x^2-3x+1}}$$
$$ =\frac{x-2}{\frac{(x-2)(x^2-3x+1)-(x^2-2x)}{x^2-3x+1}} \ \ = \  \ \frac{x-2}{\frac{x^3-3x^2+x-2x^2+6x-2-x^2+2x}{x^2-3x+1}} $$
$$ =\frac{(x-2)(x^2-3x+1)}{x^3-6x^2+9x-2} \ \ = \ \ \frac{x^3-3x^2+x-2x^2+6x-2}{x^3-6x^2+9x-2} $$
$$ =\frac{x^3-5x^2+7x-2}{x^3-6x^2+9x-2} $$
But the answer is given as $ \ \frac{x^2-3x+1}{x^2-4x+1} \ $, so I went wrong somewhere but can't see it. Any help is appreciated thanks.
 [/INST] Your answer is correct.... Sort of!
The answer you arrived at, can be simplified further to arrive at the answer you wish to achieve, since your attempt is completely valid. So, we have:
$$\frac{x^3-5x^2+7x-2}{x^3-6x^2+9x-2}$$
$$\frac{x^3-5x^2+7x-8+6}{x^3-6x^2+9x-8+6}$$
$$\frac{(x^3-8)-(5x^2-7x-6)}{(x^3-8)-(6x^2-9x-6)}$$
$$\frac{(x-2)(x^2+2x+4)-(5x^2-10x+3x-6)}{(x-2)(x^2+2x+4)-(6x^2-12x+3x-6)}$$
$$\frac{(x-2)(x^2+2x+4)-[5x(x-2)+3(x-2)]}{(x-2)(x^2+2x+4)-[6x(x-2)+3(x-2)]}$$
Factoring out $(x-2)$ from the numerator and denominator and canceling them out, we get:
$$\frac{x^2+2x+4-5x-3}{x^2+2x+4-6x-3}$$
$$\frac{x^2-3x+1}{x^2-4x+1}$$
And that's the desired answer.
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