Potrzebuje tego daje 100pkt oraz najj!!!Rozwiaz nierownosci.[tex]a) |x|\geq 7\\b) |2x|\ \textgreater.... Question from @Brszz07

Szczegółowe wyjaśnienie:

$a)\ |x|\geq7\Rightarrow x\geq7\ \vee\ x\leq-7\\\\\huge\boxed{x\in\left(-\infty,\ -7\right>\ \cup\ \left$

$b)\ |2x|>\dfrac{8}{9}\Rightarrow2x>\dfrac{8}{9}\ \vee\ 2x\dfrac{4}{9}\ \vee\ x$

$c)\ \left|x+\dfrac{1}{3}\right|\geq3\Rightarrow x+\dfrac{1}{3}\geq3\ \vee\ x+\dfrac{1}{3}\leq-3\qquad|-\dfrac{1}{3}\\\\x\geq2\dfrac{2}{3}\ \vee\ x\leq-3\dfrac{1}{3}\\\\\huge\boxed{x\in\left(-\infty,\ -3\dfrac{1}{3}\right>\ \cup\ \left$

$d)\ \left|\dfrac{2}{5}x-6\right|>1\Rightarrow\dfrac{2}{5}x-6>1\ \vee\ \dfrac{2}{5}x-67\ \vee\ \dfrac{2}{5}x\dfrac{35}{2}\ \vee\ x17,5\ \vee\ x$

2 votes Thanks 3

MiyukiSia

a) |x| ≥ 7

x ≥ 7, x ≥ 0

-x ≥ 7, x < 0

x∈ [7, + ∞>

x ≤ -7, x < 0

x∈ [7, + ∞>

x∈ < -∞, -7]

x∈ < -∞, -7] ∪ [7, + ∞>

b) |2x| > $\frac{8}{9}$

2 × |x| > $\frac{8}{9}$

2x > $\frac{8}{9}$ , x ≥ 0

2 × (-x) > $\frac{8}{9}$ , x < 0

x > $\frac{4}{9}$ , x ≥ 0

x < - $\frac{4}{9}$ , x < 0

x∈ < $\frac{4}{9}$ , + ∞ >

x∈ < - ∞, - $\frac{4}{9}$ >

x∈ < - ∞, - $\frac{4}{9}$ > ∪ < $\frac{4}{9}$ , + ∞ >

c) |x + $\frac{1}{3}$ | ≥ 3

x + $\frac{1}{3}$ ≥ 3, x + $\frac{1}{3}$ ≥ 0

- (x + $\frac{1}{3}$ ) ≥ 3, x + $\frac{1}{3}$ < 0

x ≥ $\frac{8}{3}$ , x ≥ - $\frac{1}{3}$

x ≤ - $\frac{10}{3}$ , x < - $\frac{1}{3}$

x∈ [ $\frac{8}{3}$ , + ∞ >

x∈ < - ∞, - $\frac{10}{3}$ ]

x∈ < - ∞, - $\frac{10}{3}$ ] ∪ [ $\frac{8}{3}$ , + ∞ >

d) | $\frac{2}{5}$ x - 6| > 1

$\frac{2}{5}$ x - 6 > 1, $\frac{2}{5}$ x - 6 ≥ 0

- ( $\frac{2}{5}$ x - 6 ) > 1, $\frac{2}{5}$ x - 6 < 0

x > $\frac{35}{2}$ , x ≥ 15

x < $\frac{25}{2}$ , x < 15

x∈ < $\frac{35}{2}$ , + ∞ >

x∈ < - ∞, $\frac{25}{2}$ >

x∈ < - ∞, $\frac{25}{2}$ > ∪ < $\frac{35}{2}$ , + ∞ >

Mam nadzieję, że pomogłam. Liczę na naj :)

5 votes Thanks 5

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