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Question Number 182571 by Khalmohmmad last updated on 11/Dec/22

	Answered by aleks041103 last updated on 11/Dec/22
	$((x^2 .((x^2 .((x^2 ....))^(1/3) ))^(1/3) ))^(1/3) =y⇒y=((x^2 y))^(1/3) ⇒y^3 =x^2 y⇒y=±x, but obv. y>0⇒y=∣x∣ (√(x^2 +x(√(x^2 +x(√(x^2 +x....))))))=z ⇒(√(x^2 +xz))=z ⇒z^2 =x^2 +xz⇒z^2 −xz−x^2 =0 ⇒z_(1/2) =((x±(√(x^2 +4x^2 )))/2)=((1±(√5))/2)x but z>0⇒z= { ((((1+(√5))/2)x, x>0)),((((1−(√5))/2)x, x<0)) :} ⇒5= { ((((1+(√5))/2) (x/(∣x∣)),x>0)),((((1−(√5))/2) (x/(∣x∣)),x<0)) :}= { ((((1+(√5))/2),x>0)),((undef x=0)),(((((√5)−1)/2),x<0)) :} ⇒No solution!$
	$\sqrt[3]{x^{2} . \sqrt[3]{x^{2} . \sqrt[3]{x^{2} . . . .}}} = y \Rightarrow y = \sqrt[3]{x^{2} y}$ $\Rightarrow y^{3} = x^{2} y \Rightarrow y = \pm x, b u t o b v . y > 0 \Rightarrow y =∣ x ∣$ $\sqrt{x^{2} + x \sqrt{x^{2} + x \sqrt{x^{2} + x . . . .}}} = z$ $\Rightarrow \sqrt{x^{2} + x z} = z$ $\Rightarrow z^{2} = x^{2} + x z \Rightarrow z^{2} - x z - x^{2} = 0$ $\Rightarrow z_{1 / 2} = \frac{x \pm \sqrt{x^{2} + 4 x^{2}}}{2} = \frac{1 \pm \sqrt{5}}{2} x$ $b u t z > 0 \Rightarrow z = {\begin{cases} \frac{1 + \sqrt{5}}{2} x, x > 0 \\ \frac{1 - \sqrt{5}}{2} x, x < 0 \end{cases}$ $\Rightarrow 5 = {\begin{cases} \frac{1 + \sqrt{5}}{2} \frac{x}{∣ x ∣}, x > 0 \\ \frac{1 - \sqrt{5}}{2} \frac{x}{∣ x ∣}, x < 0 \end{cases} = {\begin{cases} \frac{1 + \sqrt{5}}{2}, x > 0 \\ u n d e f x = 0 \\ \frac{\sqrt{5} - 1}{2}, x < 0 \end{cases}$ $\Rightarrow N o s o l u t i o n!$
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