• Matéria: Matemática
  • Autor: flordelins2
  • Perguntado 4 anos atrás

Resolva as inequações
a: |3x − 1| < 2
b: |3x − 1| < 2
c: |x + 3| > 1
d: |x + 1| < |2x − 1|
e: |x − 3| < x + 1
(f) |x + 1| + |x − 2| >

Respostas

respondido por: niltonjunior20oss764
1

a)

\boxed{|3x-1|&lt;2}\ \therefore\

i)\ 3x-1&lt;2\ \therefore\ 3x&lt;3\ \therefore\ x&lt;1

ii)\ 3x-1&gt;-2\ \therefore\ 3x&gt;-1\ \therefore\ x&gt;-\dfrac{1}{3}

\boxed{\mathbb{S}=\left\{x\in\mathbb{R}\ |\ -\dfrac{1}{3}&lt;x&lt;1\right\}}

b)

O mesmo que a resolução anterior.

c)

\boxed{|x+3|&gt;1}\ \therefore

i)\ x+3&gt;1\ \therefore\ x&gt;-2

ii)\ x+3&lt;-1\ \therefore\ x&lt;-4

\boxed{\mathbb{S}=\left\{x\in\mathbb{R}\ |\ x\in(-\infty,-4[\cup]-2,\infty)\right\}}

d)

\boxed{|x+1|&lt;|2x-1|}\ \therefore

i)\ x+1=2x-1\ \therefore\ x=2

ii)\ x+1=-(2x-1)\ \therefore\ x+1=-2x+1\ \therefore\ x=0

x&lt;0\ \to\ \text{p. ex.}\ x=-1\ \to\ 0&lt;3\ \to\ \text{Ok.}\\\\ 0&lt;x&lt;2\ \to\ \text{p. ex.}\ x=1\ \to\ 2\nless1\\\\ x&gt;2\ \to\ \text{p. ex.}\ x=3\ \to\ 4&lt;5\ \to\ \text{Ok.}

\boxed{\mathbb{S}=\left\{x\in\mathbb{R}\ |\ x\in(-\infty,0[\cup]2,\infty)\right\}}

e)

\boxed{|x-3|&lt;x+1}\ \therefore

i)\ x-3=-(x+1)\ \therefore x-3=-x-1\ \therefore\ x=1

x&lt;1\ \to\ \text{p. ex.}\ x=0\ \to\ 3\nless1\\\\ x=1\ \to\ 2\nless2\\\\ x&gt;1\ \to\ \text{p. ex.}\ x=2\ \to\ 1&lt;3\ \to\ \text{Ok.}

\boxed{\mathbb{S}=\left\{x\in\mathbb{R}\ |\ x\in]1,\infty)\right\}}

f)

Incompleto. Não é possível resolver.


flordelins2: Obrigado. Tem algum video ou site com uma explicação
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