Question

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17. Rezolvă, în mulțimea numerelor reale, inecuațiile:
a) |¹| ≤ 6; b) |x + 51 < 0; c) |2x + 1| ≤ 3; d) |-x+4) ≤ 0; e) |3x − 2) < 1.

18. determina mulțimile:
A= {x€R|5x-8>2x+3};B={x€R|2x +3supra 5<1};C={x€R|-4<3x+2<7};D={x€R| |x +1|< radical 5} ;E={x€R|-1<5x-6 supra 2<3}
​

Answer 1

17.

Modulul

-valoarea absoluta a numarului, definit astfel:

$|x|=\left \{ {{-x,x < 0} \atop {x,x\geq 0} \right.$

|x|<k

-k<x<k

|x|≤6

-6≤x≤6

x∈[-6,6]

|x+51|<0

Nu are solutii,

modulul

fiind un numar pozitiv

|2x+1|≤3

-3≤2x+1≤3  |-1

-4≤2x≤2  |:2

-2≤x≤1

x∈[-2,1]

|-x+4|≤0

Modulul fiind un numar pozitiv⇒ -x+4=0

x=4

|3x-2|<1

-1<3x-2<1   |+2

1<3x<3  |:3

$\frac{1}{3} < x < 1\\\\x\in (\frac{1}{3} ,1)$

18.

5x-8>2x+3

3x>11

$x > \frac{11}{3}\\\\ x\in (\frac{11}{3} , +\infty)$

$\frac{2x+3}{5} < 1\\\\\frac{2x+3}{5} -1 < 0\\\\\frac{2x+3-5}{5} < 0\\\\ 2x-2 < 0\\\\x < 1\\\\x\in (-\infty, 1)$

-4<3x+2<7 |-2

-6<3x<5 |:3

$-2 < x < \frac{5}{3} \\\\x\in (-2,\frac{5}{3} )$

|x+1|<√5

-√5<x+1<√5 |-1

-√5-1<x<√5+1

x∈(-√5-1, √5+1)

$-1 < \frac{5x-6}{2} < 3\ \ \cdot 2\\\\-2 < 5x-6 < 6\ \ \ |+6\\\\4 < 5x < 12\ \ |:5\\\\x\in(\frac{4}{5}, \frac{12}{5})$

Un alt exercitiu gasesti aici: https://brainly.ro/tema/5039359

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