Question

2-Resolva as inequações:

Answer 1

$\displaystyle \left|1-\frac{4x}{5}\right|\geq 3\iff 1-\frac{4x}{5}\geq 3 \vee 1-\frac{4x}{5}\leq-3 \\ \\ \left|1-\frac{4x}{5}\right|\geq 3\iff \frac{4x}{5}\leq -2 \vee \frac{4x}{5}\geq4 \\ \\ \left|1-\frac{4x}{5}\right|\geq 3\iff x\leq -\frac{5}{2} \vee x\geq5 \\ \\ \left|1-\frac{4x}{5}\right|\geq 3\iff \boxed{x\in \left(-\infty ; -\frac{5}{2}\right] \cup \left[5;+\infty\right)}$

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$2\ \textless \ |x+1|\ \textless \ 5\iff |x+1|\ \textgreater \ 1 \wedge |x+1|\ \textless \ 5\\ \\ 2\ \textless \ |x+1|\ \textless \ 5\iff (x+1\ \textgreater \ 1 \vee x+1\ \textless \ -1) \wedge (-5\ \textless \ x+1\ \textless \ 5)\\ \\ 2\ \textless \ |x+1|\ \textless \ 5\iff (x\ \textgreater \ 0 \vee x\ \textless \ -2) \wedge (-6\ \textless \ x\ \textless \ 4)\\ \\ 2\ \textless \ |x+1|\ \textless \ 5\iff x\in [(-\infty,-2)\cup (0,+\infty)] \cap (-6,4)\\ \\ 2\ \textless \ |x+1|\ \textless \ 5\iff \boxed{x\in (-6,-2)\cup (0,4)}$

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$1\ \textless \ |x-1|\leq 2\iff |x-1|\ \textgreater \ 1 \wedge |x-1|\leq 2\\ \\ 1\ \textless \ |x-1|\leq 2\iff (x-1\ \textgreater \ 1 \vee x-1\ \textless \ -1) \wedge (-2\leq x-1\leq 2)\\ \\ 1\ \textless \ |x-1|\leq 2\iff (x\ \textgreater \ 2 \vee x\ \textless \ 0) \wedge (-1\leq x\leq 3)\\ \\ 1\ \textless \ |x-1|\leq 2\iff x\in [(-\infty, 0)\cup (2,+\infty)] \cap [-1,3] \\ \\ 1\ \textless \ |x-1|\leq 2\iff \boxed{x\in [-1,0)\cup (2,3]}$

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$|x^2-4|\ \textless \ 3x\iff (-3x\ \textless \ x^2-4\ \textless \ 3x) \wedge x\ \textgreater \ 0\\ \\ |x^2-4|\ \textless \ 3x\iff (x^2+3x-4\ \textgreater \ 0 \wedge x^2-3x-4\ \textless \ 0) \wedge x\ \textgreater \ 0\\ \\ |x^2-4|\ \textless \ 3x\iff [(x-1)(x+4)\ \textgreater \ 0 \wedge (x+1)(x-4)\ \textless \ 0] \wedge x\ \textgreater \ 0\\ \\ |x^2-4|\ \textless \ 3x\iff \boxed{x\in (1,4)}$