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5-x-1-3-x-2-4-lt-0-

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Question Number 208076 by hardmath last updated on 04/Jun/24
(5 − ∣x∣)^(− (1/3)) (x^2 − 4) < 0
(5−∣x∣)−13(x2−4)<0
Answered by TonyCWX08 last updated on 04/Jun/24
This inequality are defined when x∈⟨−5,5⟩ Two Possible Cases Case 1: (5−∣x∣)^(−(1/3)) <0 x^2 −4>0 (1/(5−∣x∣))<0 x^2 >4 x∈⟨−∞,−5⟩∪⟨5,∞⟩ x∈⟨−∞,−2⟩∪⟨2,∞⟩ Intersection = ⟨−∞,−5⟩∪⟨5,∞⟩ Case 2 (5−∣x∣)^(−(1/3)) >0 x^2 −4<0 (1/(5−∣x∣))>0 x^2 <4 x∈⟨−5,5⟩ x∈⟨−2,2⟩ Intersection = ⟨−2,2⟩ Total Intersection= ⟨−∞,−5⟩∪⟨−2,2⟩∪⟨5,∞⟩ Defined range = x∈⟨−5,5⟩ Solution : x∈⟨−2,2⟩
Thisinequalityaredefinedwhenx∈⟨−5,5⟩TwoPossibleCasesCase1:(5−∣x∣)−13<0x2−4>015−∣x∣<0x2>4x∈⟨−∞,−5⟩∪⟨5,∞⟩x∈⟨−∞,−2⟩∪⟨2,∞⟩Intersection=⟨−∞,−5⟩∪⟨5,∞⟩Case2(5−∣x∣)−13>0x2−4<015−∣x∣>0x2<4x∈⟨−5,5⟩x∈⟨−2,2⟩Intersection=⟨−2,2⟩TotalIntersection=⟨−∞,−5⟩∪⟨−2,2⟩∪⟨5,∞⟩Definedrange=x∈⟨−5,5⟩Solution:x∈⟨−2,2⟩
Answered by lepuissantcedricjunior last updated on 04/Jun/24
x^2 −4=0 ou 5−∣x∣=0 =>x=∓2 ou { ((5+x=0)),((5−x=0)) :}=> { ((x=−5)),((x=5)) :} S_R ={−5;−2;2;5}
x2−4=0ou5−∣x∣=0=>x=∓2ou{5+x=05−x=0=>{x=−5x=5SR={−5;−2;2;5}

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