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Question-168013

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Question Number 168013 by leicianocosta last updated on 31/Mar/22
Answered by nurtani last updated on 31/Mar/22
^9 (√x)+((^9 (√x^8 ))/x)=((17)/4) ⇔ x^(1/9) +(x^(8/9) /x)=((17)/4) ⇔ x^(1/9) +x^((8/9)−1) =((17)/4) ⇔ x^(1/9) +x^(−(1/9)) =((17)/4) ⇔ x^(1/9) +(1/x^(1/9) )=((17)/4) say : x^(1/9) = β ⇒ β+(1/β) = ((17)/4) ⇒ β^2 +1=((17)/4)β × 4 ⇒ 4β^2 +4=17β ⇒ 4β^2 −17β+4=0 ⇒ (β−4)(4β−1)=0 { ((β=4 ⇒ x^(1/9) =4 ⇒ x=4^9 =(2^2 )^9 =2^(18) )),((β=(1/4)⇒ x^(1/9) =(1/4)=4^(−1) ⇒ x^(1/9) =4^(−1) ⇒x=2^(−18) )) :} ∴ x = 2^(18) ∨ x = 2^(−18)
$$\:\:\:\:\:^{\mathrm{9}} \sqrt{{x}}+\frac{\:^{\mathrm{9}} \sqrt{{x}^{\mathrm{8}} }}{{x}}=\frac{\mathrm{17}}{\mathrm{4}} \\ $$$$\Leftrightarrow\:{x}^{\frac{\mathrm{1}}{\mathrm{9}}} +\frac{{x}^{\frac{\mathrm{8}}{\mathrm{9}}} }{{x}}=\frac{\mathrm{17}}{\mathrm{4}} \\ $$$$\Leftrightarrow\:{x}^{\frac{\mathrm{1}}{\mathrm{9}}} +{x}^{\frac{\mathrm{8}}{\mathrm{9}}−\mathrm{1}} =\frac{\mathrm{17}}{\mathrm{4}} \\ $$$$\Leftrightarrow\:{x}^{\frac{\mathrm{1}}{\mathrm{9}}} +{x}^{−\frac{\mathrm{1}}{\mathrm{9}}} =\frac{\mathrm{17}}{\mathrm{4}} \\ $$$$\Leftrightarrow\:{x}^{\frac{\mathrm{1}}{\mathrm{9}}} +\frac{\mathrm{1}}{{x}^{\frac{\mathrm{1}}{\mathrm{9}}} }=\frac{\mathrm{17}}{\mathrm{4}} \\ $$$${say}\::\:{x}^{\frac{\mathrm{1}}{\mathrm{9}}} =\:\beta \\ $$$$\Rightarrow\:\beta+\frac{\mathrm{1}}{\beta}\:=\:\frac{\mathrm{17}}{\mathrm{4}}\:\Rightarrow\:\beta^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{17}}{\mathrm{4}}\beta\:\:×\:\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{4}\beta^{\mathrm{2}} +\mathrm{4}=\mathrm{17}\beta\:\Rightarrow\:\mathrm{4}\beta^{\mathrm{2}} −\mathrm{17}\beta+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\:\left(\beta−\mathrm{4}\right)\left(\mathrm{4}\beta−\mathrm{1}\right)=\mathrm{0}\:\begin{cases}{\beta=\mathrm{4}\:\Rightarrow\:{x}^{\frac{\mathrm{1}}{\mathrm{9}}} =\mathrm{4}\:\Rightarrow\:{x}=\mathrm{4}^{\mathrm{9}} =\left(\mathrm{2}^{\mathrm{2}} \right)^{\mathrm{9}} =\mathrm{2}^{\mathrm{18}} }\\{\beta=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow\:{x}^{\frac{\mathrm{1}}{\mathrm{9}}} =\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{4}^{−\mathrm{1}} \Rightarrow\:{x}^{\frac{\mathrm{1}}{\mathrm{9}}} =\mathrm{4}^{−\mathrm{1}} \Rightarrow{x}=\mathrm{2}^{−\mathrm{18}} }\end{cases} \\ $$$$\:\:\:\therefore\:\:{x}\:=\:\mathrm{2}^{\mathrm{18}} \:\:\vee\:\:{x}\:=\:\mathrm{2}^{−\mathrm{18}} \\ $$
Answered by yogamulyadi last updated on 01/Apr/22
(x)^(1/9) +(1/( (x)^(1/9) ))=4+(1/4) (x)^(1/9) =4^(±1) =2^(±2) x∈{2^(-18) , 2^(18) } . . sorry, not understand the language
$$\sqrt[{\mathrm{9}}]{{x}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{9}}]{{x}}}=\mathrm{4}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\sqrt[{\mathrm{9}}]{{x}}=\mathrm{4}^{\pm\mathrm{1}} =\mathrm{2}^{\pm\mathrm{2}} \\ $$$${x}\in\left\{\mathrm{2}^{-\mathrm{18}} ,\:\mathrm{2}^{\mathrm{18}} \right\} \\ $$$$. \\ $$$$. \\ $$$${sorry},\:{not}\:{understand}\:{the}\:{language} \\ $$

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