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Question Number 168013 by leicianocosta last updated on 31/Mar/22

	Answered by nurtani last updated on 31/Mar/22
	$^9 (√x)+((^9 (√x^8 ))/x)=((17)/4) ⇔ x^(1/9) +(x^(8/9) /x)=((17)/4) ⇔ x^(1/9) +x^((8/9)−1) =((17)/4) ⇔ x^(1/9) +x^(−(1/9)) =((17)/4) ⇔ x^(1/9) +(1/x^(1/9) )=((17)/4) say : x^(1/9) = β ⇒ β+(1/β) = ((17)/4) ⇒ β^2 +1=((17)/4)β × 4 ⇒ 4β^2 +4=17β ⇒ 4β^2 −17β+4=0 ⇒ (β−4)(4β−1)=0 { ((β=4 ⇒ x^(1/9) =4 ⇒ x=4^9 =(2^2 )^9 =2^(18) )),((β=(1/4)⇒ x^(1/9) =(1/4)=4^(−1) ⇒ x^(1/9) =4^(−1) ⇒x=2^(−18) )) :} ∴ x = 2^(18) ∨ x = 2^(−18)$
	$^{9} \sqrt{x} + \frac{^{9} \sqrt{x^{8}}}{x} = \frac{17}{4}$ $\Leftrightarrow x^{\frac{1}{9}} + \frac{x^{\frac{8}{9}}}{x} = \frac{17}{4}$ $\Leftrightarrow x^{\frac{1}{9}} + x^{\frac{8}{9} - 1} = \frac{17}{4}$ $\Leftrightarrow x^{\frac{1}{9}} + x^{- \frac{1}{9}} = \frac{17}{4}$ $\Leftrightarrow x^{\frac{1}{9}} + \frac{1}{x^{\frac{1}{9}}} = \frac{17}{4}$ $s a y : x^{\frac{1}{9}} = β$ $\Rightarrow β + \frac{1}{β} = \frac{17}{4} \Rightarrow β^{2} + 1 = \frac{17}{4} β \times 4$ $\Rightarrow 4 β^{2} + 4 = 17 β \Rightarrow 4 β^{2} - 17 β + 4 = 0$ $\Rightarrow (β - 4) (4 β - 1) = 0 {\begin{cases} β = 4 \Rightarrow x^{\frac{1}{9}} = 4 \Rightarrow x = 4^{9} = {(2^{2})}^{9} = 2^{18} \\ β = \frac{1}{4} \Rightarrow x^{\frac{1}{9}} = \frac{1}{4} = 4^{- 1} \Rightarrow x^{\frac{1}{9}} = 4^{- 1} \Rightarrow x = 2^{- 18} \end{cases}$ $∴ x = 2^{18} \lor x = 2^{- 18}$
	Answered by yogamulyadi last updated on 01/Apr/22
	$(x)^(1/9) +(1/( (x)^(1/9) ))=4+(1/4) (x)^(1/9) =4^(±1) =2^(±2) x∈{2^(-18) , 2^(18) } . . sorry, not understand the language$
	$\sqrt[9]{x} + \frac{1}{\sqrt[9]{x}} = 4 + \frac{1}{4}$ $\sqrt[9]{x} = 4^{\pm 1} = 2^{\pm 2}$ $x \in {2^{- 18}, 2^{18}}$ $.$ $.$ $s o r r y, n o t u n d e r s t a n d t h e l a n g u a g e$
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