Whakaoti i te x^2+5/x^2=1

$${ x }^{ 2 } + \frac{ 5 }{ { x }^{ 2 } } =1$$

Whakaoti mō x (complex solution)

$$x=\sqrt[4]{5}e^{\frac{\arctan(\sqrt{19})i}{2}}\approx 1.169629851+0.931683417i$$

$$x=\sqrt[4]{5}e^{\frac{\arctan(\sqrt{19})i+2\pi i}{2}}\approx -1.169629851-0.931683417i$$

$$x=\sqrt[4]{5}e^{-\frac{\arctan(\sqrt{19})i}{2}}\approx 1.169629851-0.931683417i$$

$$x=\sqrt[4]{5}e^{\frac{-\arctan(\sqrt{19})i+2\pi i}{2}}\approx -1.169629851+0.931683417i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x^{2}.

$$x^{2}x^{2}+5=x^{2}$$

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 2 me te 2 kia riro ai te 4.

$$x^{4}+5=x^{2}$$

Tangohia te x^{2} mai i ngā taha e rua.

$$x^{4}+5-x^{2}=0$$

Whakakapia te t mō te x^{2}.

$$t^{2}-t+5=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 1 mō te a, te -1 mō te b, me te 5 mō te c i te ture pūrua.

$$t=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 5}}{2}$$

Mahia ngā tātaitai.

$$t=\frac{1±\sqrt{-19}}{2}$$

Whakaotia te whārite t=\frac{1±\sqrt{-19}}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$t=\frac{1+\sqrt{19}i}{2} t=\frac{-\sqrt{19}i+1}{2}$$

I te mea ko x=t^{2}, ka riro ngā otinga mā te arotake i te x=±\sqrt{t} mō ia t.

$$x=\sqrt[4]{5}e^{\frac{\arctan(\sqrt{19})i+2\pi i}{2}} x=\sqrt[4]{5}e^{\frac{\arctan(\sqrt{19})i}{2}} x=\sqrt[4]{5}e^{-\frac{\arctan(\sqrt{19})i}{2}} x=\sqrt[4]{5}e^{\frac{-\arctan(\sqrt{19})i+2\pi i}{2}}$$

Tē taea kia ōrite te tāupe x ki 0.

$$x=\sqrt[4]{5}e^{\frac{-\arctan(\sqrt{19})i+2\pi i}{2}}\text{, }x\neq 0 x=\sqrt[4]{5}e^{-\frac{\arctan(\sqrt{19})i}{2}}\text{, }x\neq 0 x=\sqrt[4]{5}e^{\frac{\arctan(\sqrt{19})i}{2}}\text{, }x\neq 0 x=\sqrt[4]{5}e^{\frac{\arctan(\sqrt{19})i+2\pi i}{2}}\text{, }x\neq 0$$