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Question Number 124988 by bemath last updated on 07/Dec/20

Commented by bemath last updated on 07/Dec/20

x = ((−1±(√(1−4)))/2) = ((−1±i(√3))/2) x_1 = −(1/2)+((i(√3))/2) ⇒x_1 = cos (((2π)/3))+ isin (((2π)/3)) x_1 ^(2007) = cos (((4014π)/3))+isin (((4014π)/3)) x_1 ^(−2007) = cos (((4014π)/3))−isin (((4014π)/3)) x^(2007) +(1/x^(2007) ) = 2cos (((4014π)/3)) = 2cos (1338π) = 2cos (0) = 2×(1)=2

$${x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}}}{\mathrm{2}}\:=\:\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}+\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow{x}_{\mathrm{1}} =\:\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)+\:{i}\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2007}} \:=\:\mathrm{cos}\:\left(\frac{\mathrm{4014}\pi}{\mathrm{3}}\right)+{i}\mathrm{sin}\:\left(\frac{\mathrm{4014}\pi}{\mathrm{3}}\right) \\ $$$${x}_{\mathrm{1}} ^{−\mathrm{2007}} \:=\:\mathrm{cos}\:\left(\frac{\mathrm{4014}\pi}{\mathrm{3}}\right)−{i}\mathrm{sin}\:\left(\frac{\mathrm{4014}\pi}{\mathrm{3}}\right) \\ $$$${x}^{\mathrm{2007}} +\frac{\mathrm{1}}{{x}^{\mathrm{2007}} }\:=\:\mathrm{2cos}\:\left(\frac{\mathrm{4014}\pi}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2cos}\:\left(\mathrm{1338}\pi\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2cos}\:\left(\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}×\left(\mathrm{1}\right)=\mathrm{2} \\ $$

Commented by floor(10²Eta[1]) last updated on 07/Dec/20

desnecessary long result

$$\mathrm{desnecessary}\:\mathrm{long}\:\mathrm{result} \\ $$

Commented by bemath last updated on 08/Dec/20

i like a long thing

$${i}\:{like}\:{a}\:{long}\:{thing}\: \\ $$

Answered by floor(10²Eta[1]) last updated on 07/Dec/20

x^2 =−x−1 x^3 =−x^2 −x=x+1−x=1 x^3 =1 x^(2007) =(x^3 )^(669) =1 x^(2007) +(1/x^(2007) )=1+1=2

$$\mathrm{x}^{\mathrm{2}} =−\mathrm{x}−\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{3}} =−\mathrm{x}^{\mathrm{2}} −\mathrm{x}=\mathrm{x}+\mathrm{1}−\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{3}} =\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{2007}} =\left(\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{669}} =\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{2007}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2007}} }=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$

Commented by talminator2856791 last updated on 07/Dec/20

please explain −x^2 −x = x+1−x

$$\:\mathrm{please}\:\mathrm{explain}\:−{x}^{\mathrm{2}} −{x}\:=\:{x}+\mathrm{1}−{x} \\ $$

Commented by floor(10²Eta[1]) last updated on 07/Dec/20

x^2 =−x−1

$$\mathrm{x}^{\mathrm{2}} =−\mathrm{x}−\mathrm{1} \\ $$

Commented by talminator2856791 last updated on 07/Dec/20

ooooooooooooo wow

$$\:\mathrm{ooooooooooooo}\:\mathrm{wow} \\ $$

Commented by talminator2856791 last updated on 07/Dec/20

can you please look at my answer for post 124978

$$\:\mathrm{can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{look}\:\mathrm{at}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{for}\:\mathrm{post}\:\mathrm{124978} \\ $$

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