Question

Determine el conjunto solución de ││2x-1│-4│≤2

Answer 1

$\left||2x-1|-4\right|\leq 2\iff -2\leq |2x-1|-4\leq 2\\ \\ \left||2x-1|-4\right|\leq 2\iff 2\leq |2x-1|\leq 6\\ \\ \left||2x-1|-4\right|\leq 2\iff \left(|2x-1|\geq 2\wedge|2x-1|\leq 6\right)\\ \\ \left||2x-1|-4\right|\leq 2\iff \left[(2x-1\leq -2 \vee 2x-1\geq 2)\wedge -6\leq 2x-1\leq 6\right]$

$\left||2x-1|-4\right|\leq 2\iff \left[(2x\leq -1 \vee 2x\geq 3)\wedge -5\leq 2x\leq 7\right]\\ \\ \left||2x-1|-4\right|\leq 2\iff \left[\left(x\leq -\dfrac{1}{2} \vee x\geq \dfrac{3}{2}\right)\wedge -\dfrac{5}{2}\leq x\leq \dfrac{7}{2}\right]$


$\left||2x-1|-4\right|\leq 2\iff x\in \left\{\left[\left(-\infty,-\dfrac{1}{2}\right]\cup \left[\dfrac{3}{2},+\infty\right)\right]\wedge \left[-\dfrac{5}{2},\dfrac{7}{2}\right]\right\}\\ \\ \\ \boxed{\boxed{\left||2x-1|-4\right|\leq 2\iff x\in\left[-\dfrac{5}{2},-\dfrac{1}{{2}\right]\cup\left[\dfrac{3}{2},\dfrac{7}{2}\right]}}}$