I am getting stuck at the start of the expression. I cant understand how to properly interpret the absolute brackets and how to get to three conditions x < 1, 1 < x < 2 and x > 2. $$(((x^2|2-x|)/(x-2))+2x-4-((2x|x-1|)/(x-1)))/|x-2|$$
This is how I start, is it correct? What am I doing wrong?
$|2-x|={2-x > 0 , -x > -2 , x < 2, and -(2-x) < 0 , x < 2 } $ $|x-2|={x-2 > 0 , x > 2, and -(x-2) < 0 , -x < -2, x > 2 } $ $|x-1|={x-1 > 0 , x > 1, and -(x-1) < 0 , -x < -1, x > 1 } $
Also, this isn't part of an equation, I only need to reduce this expression as far as it's possible.
Corrected; the original answer was for a misreading of the nested parentheses.
It’s helpful to rewrite the expression as
$$\frac{x^2|2-x|}{(x-2)|x-2|}+\frac{2(x-2)}{|x-2|}-\frac{2x|x-1|}{(x-1)|x-2|}\;.$$
Clearly this is undefined where a denominator is $0$, i.e., at $x=1$ and $x=2$. This naturally breaks the real line into three pieces, $x<1$, $1<x<2$, and $x>2$. And since we’re not going to be considering $x=2$ anyway, we can take advantage of the fact that $|2-x|=|x-2|$ to simplify the first term to $\frac{x^2}{x-2}$, so that we have
$$\frac{x^2}{x-2}+\frac{2(x-2)}{|x-2|}-\frac{2x|x-1|}{(x-1)|x-2|}\;.$$
If $x>2$, then $|x-2|=x-2$, so the second term simplifies to $2$. Moreover, in that case $x>1$, so $|x-1|=x-1$, and the last term simplifes to $\frac{2x}{x-2}$, making the whole thing
$$\frac{x^2}{x-2}+2-\frac{2x}{x-2}=\frac{x^2-4}{x-2}=\frac{(x+2)(x-2)}{x+2}=x+2$$
when $x>2$. What if $1<x<2$? Then $|x-2|=2-x=-(x-2)$, so the second term simplifies to $-2$ and the last term to $+\frac{2x}{x-2}$, making the whole thing
$$\frac{x^2}{x-2}-2+\frac{2x}{x-2}=\frac{x^2+4}{x-2}$$
when $1<x<2$.
Finally, if $x<1$, then $|x-1|=1-x=-(x-1)$, and $|x-2|=2-x=-(x-2)$, so the whole thing simplifies to
$$\frac{x^2}{x-2}-2-\frac{2x}{x-2}=\frac{x^2-4x+4}{x-2}=\frac{(x-2)^2}{x-2}=x-2$$
when $x<1$.