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Question Number 208076 by hardmath last updated on 04/Jun/24
(5 − ∣x∣)^(− (1/3)) (x^2 − 4) < 0
$$\left(\mathrm{5}\:−\:\mid\mathrm{x}\mid\right)^{−\:\frac{\mathrm{1}}{\mathrm{3}}} \:\left(\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{4}\right)\:<\:\mathrm{0} \\ $$
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⟩
$${This}\:{inequality}\:{are}\:{defined}\:{when}\:{x}\in\langle−\mathrm{5},\mathrm{5}\rangle \\ $$$${Two}\:{Possible}\:{Cases} \\ $$$${Case}\:\mathrm{1}: \\ $$$$\left(\mathrm{5}−\mid{x}\mid\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} <\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}>\mathrm{0} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{5}−\mid{x}\mid}<\mathrm{0} \\ $$$${x}^{\mathrm{2}} >\mathrm{4} \\ $$$$ \\ $$$${x}\in\langle−\infty,−\mathrm{5}\rangle\cup\langle\mathrm{5},\infty\rangle \\ $$$${x}\in\langle−\infty,−\mathrm{2}\rangle\cup\langle\mathrm{2},\infty\rangle \\ $$$$ \\ $$$${Intersection}\:=\:\langle−\infty,−\mathrm{5}\rangle\cup\langle\mathrm{5},\infty\rangle \\ $$$$ \\ $$$${Case}\:\mathrm{2} \\ $$$$\left(\mathrm{5}−\mid{x}\mid\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} >\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}<\mathrm{0} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{5}−\mid{x}\mid}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} <\mathrm{4} \\ $$$$ \\ $$$${x}\in\langle−\mathrm{5},\mathrm{5}\rangle \\ $$$${x}\in\langle−\mathrm{2},\mathrm{2}\rangle \\ $$$$ \\ $$$${Intersection}\:=\:\langle−\mathrm{2},\mathrm{2}\rangle \\ $$$$ \\ $$$${Total}\:{Intersection}=\:\langle−\infty,−\mathrm{5}\rangle\cup\langle−\mathrm{2},\mathrm{2}\rangle\cup\langle\mathrm{5},\infty\rangle \\ $$$${Defined}\:{range}\:=\:{x}\in\langle−\mathrm{5},\mathrm{5}\rangle \\ $$$${Solution}\::\:{x}\in\langle−\mathrm{2},\mathrm{2}\rangle \\ $$
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}
$$\:\:\:\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{4}=\mathrm{0}\:\boldsymbol{{ou}}\:\mathrm{5}−\mid\boldsymbol{{x}}\mid=\mathrm{0} \\ $$$$=>\boldsymbol{{x}}=\mp\mathrm{2}\:\boldsymbol{{ou}}\:\begin{cases}{\mathrm{5}+\boldsymbol{{x}}=\mathrm{0}}\\{\mathrm{5}−\boldsymbol{{x}}=\mathrm{0}}\end{cases}=>\begin{cases}{\boldsymbol{{x}}=−\mathrm{5}}\\{\boldsymbol{{x}}=\mathrm{5}}\end{cases} \\ $$$$\boldsymbol{{S}}_{\mathbb{R}} =\left\{−\mathrm{5};−\mathrm{2};\mathrm{2};\mathrm{5}\right\} \\ $$

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