Question

va rog foarte mult este super super urgent va roggg​

Answer 1

Stabilesc conditiile de exista, tot patru la numar

I. 3x-5 ≥ 0 ; x+1 ≥ 0 ⇒ x ≥ 5/3; x ≥ -1 → x ∈ [5/3, infinit)

II. 3x-5 ≥ 0; x+1 <0 ⇒ x ≥ 5/3; x < -1 → x ∈ (-1, 5/3]

III. 3x-5 < 0; x+1 ≥ 0 ⇒ x < 5/3; x ≥ -1 → x ∈ ∅

IV. 3x-5 <0; x+1< 0 ⇒ x < 5/3; x < -1 → x ∈ (-infinit -1)

|3x-5 | -|x+1| = 2  Exista patru moduri de rezolvare

I. 3x-5 - (x+1) = 2

2x-6 = 2⇒ 2x = 8 ⇒ x = 4

II. 3x-5 - (-(x+1)) = 2

3x-5 -(-x-1) = 2 ⇒ 3x-5+x+1 = 2 ⇒ 4x -4 = 2 ⇒ 4x = 6 ⇒ x = 6/4 = 3/2

III. -(3x-5) -(x+1) = 2

-3x+5 -x-1 = 2 ⇒ -4x+4 = 2 ⇒ -4x = -2 ⇒ x = 1/2

IV. -(3x-5)-(-(x+1)) = 2

-3x+5 +x+1 = 2 ⇒ -2x+6= 2 ⇒ -2x = -4 ⇒ x =2

I. x =4 apartine intervalului

II. x = 3/2 nu apartine intervalului

III. x = 1/2 apartine intervalului

IV. x = 2 nu apartine intervalului

S={1/2, 4}

Answer 2

$\it \rule{50}{0.8}\Big|-\infty\rule{70}{0.8}\ -1\rule{50}{0.8}5/3\rule{60}{0.8}\ \infty\\ |3x-5|\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3x+5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big| \ \ \ \ \ \ \ 3x-5\\ \rule{50}{0.8}\Big|\rule{230}{0.8}\\|x+1| \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -x-1\ \ \ \ \ \ \ \ \ \ \ \ \ \Big|\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x+1\\ \rule{50}{0.8}\Big|\rule{230}{0.8}$

$\it I)\ x\in(-\infty,\ -1),\ \ Ecua\c{\it t}ia\ devine:\\ \\ \Rightarrow -3x+5+x+1=2 \Rightarrow -2x+6=2 \Rightarrow 6-2=2x \Rightarrow \\ \\ \Rightarrow 4=2x|_{:2} \Rightarrow 2=x \Rightarrow x=2\not\in(-\infty,\ -1)$

$\it II)\ x\in\Big[-1,\ \ \dfrac{5}{3}\Big]\ \Rightarrow Ecua\c{\it t}ia\ devine:\\ \\ -3x+5-x-1=2 \Rightarrow -4x+4=2 \Rightarrow 4-2=4x \Rightarrow 2=4x \Rightarrow \\ \\ 4x=2|_{:2} \Rightarrow 2x=1 \Rightarrow x=\dfrac{1}{2}\in[-1,\ \dfrac{5}{3}] \Rightarrow S_1=\Big\{\dfrac{1}{2}\Big\}$

$\it x\in\Big(\dfrac{5}{3},\ \infty\Big) \Rightarrow Ecua\c{\it t}ia\ devine:\\ \\ 3x-5-x-1=2 \Rightarrow 2x-6=2|_{+6} \Rightarrow 2x=8 \Rightarrow x=4\in\Big(\dfrac{5}{3},\ \ \infty\Big) \Rightarrow \\ \\ \Rightarrow S_2=\{4\}$

$\it S=\Big\{\dfrac{1}{2},\ 4\Big\}$