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

((2+(√5)))^(1/3) + ((2−(√5)))^(1/3) =? ((9^x −5.12^x + 4^(2x+1) )/(log _2 (6x^2 −11x+4))) ≤ 0

$$\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:=? \\ $$$$\:\frac{\mathrm{9}^{\mathrm{x}} −\mathrm{5}.\mathrm{12}^{\mathrm{x}} +\:\mathrm{4}^{\mathrm{2x}+\mathrm{1}} }{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{6x}^{\mathrm{2}} −\mathrm{11x}+\mathrm{4}\right)}\:\leqslant\:\mathrm{0}\: \\ $$

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)

$$\left(\mathrm{1}\right)\:\left(\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:\right)^{\mathrm{3}} \:=\:\mathrm{t}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\mathrm{4}\:+\:\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{4}−\mathrm{5}}\:\left(\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:\right)=\:\mathrm{t}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\mathrm{4}\:−\mathrm{3t}\:=\:\mathrm{t}^{\mathrm{3}} \:;\:\mathrm{t}^{\mathrm{3}} +\mathrm{3t}−\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\left(\mathrm{t}−\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} +\mathrm{t}+\mathrm{4}\right)=\mathrm{0}\:\rightarrow\begin{cases}{\mathrm{t}=\mathrm{1}}\\{\mathrm{t}^{\mathrm{2}} +\mathrm{t}+\mathrm{4}=\mathrm{0}\:\rightarrow\Delta<\mathrm{0}}\end{cases} \\ $$$$\:\:\:\:\:\mathrm{t}=\mathrm{1}=\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\: \\ $$$$\:\:\:\:\: \\ $$

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