Question

Determine o conjunto dos valores de x que satisfazem o sistema de inequações:

x² - 4x + 3 > 0
x² - 2x ≤ 0

Answer 1

Olá Brubs.


Organizando e resolvendo a equação:


$\begin{cases}{\mathsf{x^2 - 4x + 3 \ \textgreater \ 0}\\\mathsf{x^2 - 2x \leq 0}\end{cases}\\\\\\\\\mathsf{x^2-4x+3\ \textgreater \ 0}\\\\\\\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=(-4)^2-4\cdot1\cdot3}\\\mathsf{\Delta=16-12}\\\mathsf{\Delta=4}$

$\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\\mathsf{x^+=\dfrac{-(-4)+\sqrt{4}}{2\cdot1}\qquad\qquad\qquad\qquad x^-=\dfrac{(-4)-\sqrt{4}}{2\cdot1}}\\\\\\\mathsf{x^+=\dfrac{4+2}{2}\qquad\qquad\qquad\qquad\qquad~~~x^-=\dfrac{4-2}{2}}\\\\\\\mathsf{x^+=\dfrac{6}{2}\qquad\qquad\qquad\qquad\qquad\qquad ~~x^-=\dfrac{2}{2}}\\\\\\\boxed{\mathsf{x^+=3}}\qquad\qquad\qquad\qquad\qquad\qquad \boxed{\mathsf{x^-=1}}\\\\\\\\\mathsf{(x-3)\cdot(x-1)=x^2-4x+3}\\\\\\\mathsf{x-3~\Rightarrow~x=3}$

$\mathsf{x-1~\Rightarrow~x=1}$

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$\mathsf{x^2-2x \leq 0}\\\\\mathsf{x\cdot(x-2) \leq 0}\\\\\mathsf{x \leq 0}\\\\\mathsf{x-2 \leq 0}\\\mathsf{x \leq 2}$


Fazendo a intersecção dos dois sistemas:


$\begin{cases}\mathsf{(x-1)\qquad\qquad\qquad {_\underline{~--}} {_\underline{----}}\underset1\circ}}{_\underline{~+++++++++++++}}_\blacktriangleright}}}}}\\\\\mathsf{(x-3)\qquad\qquad\qquad {_\underline{~-------------}}}\underset3\circ{_\underline{++++++}}_\blacktriangleright}}}\\\\\mathsf{(x-1)\cdot(x-3)\qquad {_\underline{~++++++}}\underset1\circ{_\underline{------~}}\underset3\circ{_\underline{+++++}}_\blacktriangleright}}}\end{cases}$


$\mathsf{(x-1)\cdot(x-3)\ \textgreater \ 0}\\\\\\\mathsf{S:\{x\in\mathbb{R}:x\ \textless \ 1~ou~x\ \textgreater \ 3\}}$

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$\begin{cases}\mathsf{x\qquad\qquad\qquad\qquad~{_\underline{~------~}}\underset0\bullet{_\underline{++++++++++++++}}_\blacktriangleright}}}}}\\\\\mathsf{(x-2)\qquad\qquad\qquad {_\underline{~------------~}}}\underset2\bullet{_\underline{~+++++++}}}_\blacktriangleright}}}}}\\\\\mathsf{x\cdot(x-2)~~\qquad\qquad {_\underline{~+++++++~}}\underset0\bullet}{_\underline{~----~}}\underset2\bullet{_\underline{~+++++++}}_\blacktriangleright}}}}\end$


$\mathsf{x\cdot(x-2) \leq 0}}\\\\\\\mathsf{S:\{x\in\mathbb{R}: 0 \leq x \leq 2~\}}$

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$\begin{cases}\mathsf{(x-1)\cdot(x-3)\qquad{_\underline{.................}}\underset1\circ{_\underline{\qquad~~~ ~}}}\underset3\circ{_\underline{.................}}}_\blacktriangleright}}}}}\\\\\mathsf{x\cdot(x-2)\qquad\qquad{_\underline{~~~~}}\underset0\bullet{_\underline{..................}}}\underset2\bullet{_\underline{\qquad\qquad~\qquad}}}_\blacktriangleright}}}}}\end{cases}$

$\mathsf{\cap \qquad\qquad~~~~~~\qquad{_\underline{~~~~}}\underset0\bullet{_\underline{..........}}\underset1\circ\underline{\qquad\qquad\qquad\qquad~}} {_\blacktriangleright}}}$


Portanto, a solução do sistema é:


$\boxed{\boxed{\mathsf{S:\{x\in\mathbb{R}:0 \geq x \ \textgreater \ 1\}}}}$


Bons estudos :^) !


Dúvidas? comente.