Order of calculation about nested absolute values In case of nested absolute values, is there a difference between working first on the outer abs value and the opposite?
$$ \frac{|2x^2-3|x|+1|}{|x|+1} = 1 $$
Should I consider first the inner one?
My real solutions for this equation are $\left \{0, \pm 2\right \}$, and I've found those working first with the inner one.
 A: As long as you do it correctly, you can do whatever way you want.
I would simply do it as follows.
Let $y=|x|$. Then
$$
2y^2-3y+1=y+1,\quad y\ge 0\tag{1}
$$
or
$$
2y^2-3y+1=-(y+1),\quad y\ge 0\tag{2}
$$
Now solve (1) and (2); you can then easily find solutions to (1) and find that (2) has no solution.
A: Yet another solution.
$$\begin{align}
&|2x^2-3|x|+1|=|x|+1\\ 
\iff\quad& (2|x|-1)\cdot(|x|-1)=|x|-1+2\\
\iff\quad &(|x|-1)(2|x|-1-1)=2\\
\iff\quad & (|x|-1)^2=1\\
\iff\quad &|x|-1=\pm1\\
\iff\quad &x\in\{0,\pm2\}
\end{align}
$$
A: It is generally easier to start from the inside and then go outside. The usual procedure would be to make 2 cases: $x\geq0$ and $x<0$ which will give you:
$$\frac{|2x^2-3x+1|}{x+1} = 1 ~ \text{for} ~x \geq 0 $$
$$\frac{|2x^2+3x+1|}{-x+1} = 1 ~ \text{for} ~x < 0 $$
Now you can solve these two equations separately by further dividing each case into sub-cases of $2x^2+3x+1 \geq 0$  and $2x^2+3x+1<0$
Also note that for this particular question, $2x^2+3x+1 \geq 0 ~\forall ~x \geq0$ so, it reduces a case.
