Question

Calculează
$\sqrt{ {x}^{2} + 6x + 9 } \geqslant 2$

va rog
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Answer 1

Răspuns: $\bf x \in (-\infty, -5]\cup [-1, +\infty)$

Explicație pas cu pas:

Salutare!

$\bf \sqrt{x^{2} +6x+9} \geq 2$

$\bf (\sqrt{x^{2} +6x+9} )^{2}\geq (2)^{2}$

$\bf x^{2} +6x+9\geq 4$

$\bf x^{2} +2\cdot 3 \cdot x +3^{2}\geq 4$

$\bf (x+ 3)^{2}\geq 2^{2}$

$\bf |x+ 3|\geq 2 \implies x+3\geq 2\implies x \geq -1$

                         $\bf x+3\geq 0\implies x \geq -3$

$\bf |x+ 3|\geq 2 \implies -(x+3)\geq 2\implies x \leq -5$

                         $\bf x+3<0\implies x < -3$

$\bf x\geq -1; x\geq -3 \implies x\in [-1, +\infty)$

$\bf x\leq -5; x<-3 \implies x\in (-\infty, -5]$

$\bf x \in (-\infty, -5]\cup [-1, +\infty)$

Answer 2

Răspuns:

x€(- infinit,-5]∪[-1,+infinit)

Explicație pas cu pas:

Salutare ! ᕕ( ᐛ )ᕗ

$\sqrt{x {}^{2} + 6x + 9 } \geqslant 2$

$x \: € \: R$

$x {}^{2} + 6x + 9 \geqslant 4$

$(x + 3) {}^{2} \geqslant 4$

$|x + 3| \geqslant 2$

$x + 3 \geqslant 2 \: ; \: x + 3 \geqslant 0$

$- (x + 3) \geqslant 2 \: ; \: x + 3 < 0$

$x \geqslant - 1 \: ; \: x \geqslant - 3$

$x \leqslant - 5 \: ; \: x < - 3$

x€[-1,+infinit)

x€(- infinit,-5]

x€(- infinit,-5]∪[-1,+infinit)

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