Question

(a) Se x > 1 determine o valor da expressão 1 - |x-|1-x|| eliminando os módulos.
(b) Sabendo que x < 0, elimine os módulos de |x-|x-1||.

Answer 1

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(a) Se $\mathsf{x>1},$ então

$\mathsf{|1-x|=|x-1|=x-1}\\\\ \mathsf{\Rightarrow~~\big|x-|1-x|\big|=\big|x-(x-1)\big|=\big|1\big|=1}\\\\ \mathsf{\Rightarrow~~1-\big|x-|1-x|\big|=1-1=0\qquad\quad\checkmark}$


$\therefore~~\mathsf{1-\big|x-|1-x|\big|=0\qquad para~x>1.}$

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(b) Se $\mathsf{x<0},$ então

$\mathsf{x<0<1~~\Rightarrow~~x<1~~\Rightarrow~~|x-1|=1-x}\\\\ \mathsf{\Rightarrow~~\big|x-|x-1|\big|=\big|x-(1-x)\big|=\big|2x-1\big|\qquad(i)}\\\\\\ \mathsf{\big|2x-1\big|}=\left\{\! \begin{array}{rl} \mathsf{2x-1,}&\mathsf{se~2x-1\ge 0}\\\\ \mathsf{-(2x-1),}&\mathsf{se~2x-1<0} \end{array} \right.\qquad\quad\mathsf{~e~~x<0}\\\\\\\\ \mathsf{\big|2x-1\big|}=\left\{\! \begin{array}{rl} \mathsf{2x-1,}&\mathsf{se~x\ge \frac{1}{2}}\\\\ \mathsf{1-2x,}&\mathsf{se~x<\frac{1}{2}} \end{array} \right.\qquad\quad\mathsf{~e~~x<0}$


$\mathsf{\big|2x-1\big|=1-2x,}$    pois $\mathsf{x<0<\frac{1}{2}.}$


Por fim, temos que

$\mathsf{\big|x-|x-1|\big|=1-2x\qquad para~x<0.}$


Bons estudos! :-)