((2+(√5)))^(1/3) + ((2−(√5)))^(1/3) =? ((9^x −5.12^x + 4^(2x+1) )/(log


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Question Number 129785 by bramlexs22 last updated on 19/Jan/21

$\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} = ?$ $\frac{9^{x} - 5 . 12^{x} + 4^{2 x + 1}}{\log_{2} ({6 x}^{2} - 11 x + 4)} ⩽ 0$
	Answered by EDWIN88 last updated on 19/Jan/21
	$(1) ( ((2+(√5)))^(1/3) +((2−(√5)))^(1/3) )^3 = t^3 4 + 3((4−5))^(1/3) ( ((2+(√5)))^(1/3) + ((2−(√5)))^(1/3) )= t^3 4 −3t = t^3 ; t^3 +3t−4=0 (t−1)(t^2 +t+4)=0 → { ((t=1)),((t^2 +t+4=0 →Δ<0)) :} t=1=((2+(√5)))^(1/3) + ((2−(√5)))^(1/3)$
	$(1) {(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}})}^{3} = t^{3}$ $4 + 3 \sqrt[3]{4 - 5} (\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}) = t^{3}$ $4 - 3 t = t^{3}; t^{3} + 3 t - 4 = 0$ $(t - 1) (t^{2} + t + 4) = 0 \to {\begin{cases} t = 1 \\ t^{2} + t + 4 = 0 \to Δ < 0 \end{cases}$ $t = 1 = \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$
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