Whakaoti i te 2x+1+1/x^2=0

$$2 x + 1 + \frac { 1 } { x ^ { 2 } } = 0$$

Whakaoti mō x (complex solution)

$$x=\frac{1+\sqrt{7}i}{4}\approx 0.25+0.661437828i$$

$$x=\frac{-\sqrt{7}i+1}{4}\approx 0.25-0.661437828i$$

$$x=-1$$

Ngā Upane Otinga

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}.

$$2xx^{2}+x^{2}+1=0$$

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 1 me te 2 kia riro ai te 3.

$$2x^{3}+x^{2}+1=0$$

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau 1, ā, ka wehea e q te whakarea arahanga 2. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{1}{2},±1$$

Kimihia tētahi pūtake pērā mā te whakamātau i ngā uara tau tōpū katoa, e tīmata ana i te mea iti rawa mā te uara pū. Mēnā kāore he pūtake tau tōpū e kitea, whakamātauria ngā hautanga.

$$x=-1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 2x^{3}+x^{2}+1 ki te x+1, kia riro ko 2x^{2}-x+1. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$2x^{2}-x+1=0$$

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 2 mō te a, te -1 mō te b, me te 1 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$x=\frac{1±\sqrt{-7}}{4}$$

Whakaotia te whārite 2x^{2}-x+1=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=\frac{-\sqrt{7}i+1}{4} x=\frac{1+\sqrt{7}i}{4}$$

Rārangitia ngā otinga katoa i kitea.

$$x=-1 x=\frac{-\sqrt{7}i+1}{4} x=\frac{1+\sqrt{7}i}{4}$$

Whakaoti mō x

$$x=-1$$

Ngā Upane Otinga

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}.

$$2xx^{2}+x^{2}+1=0$$

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 1 me te 2 kia riro ai te 3.

$$2x^{3}+x^{2}+1=0$$

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau 1, ā, ka wehea e q te whakarea arahanga 2. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{1}{2},±1$$

Kimihia tētahi pūtake pērā mā te whakamātau i ngā uara tau tōpū katoa, e tīmata ana i te mea iti rawa mā te uara pū. Mēnā kāore he pūtake tau tōpū e kitea, whakamātauria ngā hautanga.

$$x=-1$$

Mā te whakatakotoranga Tauwehe, he tauwehe te x-k o te pūrau mō ia pūtake k. Whakawehea te 2x^{3}+x^{2}+1 ki te x+1, kia riro ko 2x^{2}-x+1. Whakaotihia te whārite ina ōrite te hua ki te 0.

$$2x^{2}-x+1=0$$

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 2 mō te a, te -1 mō te b, me te 1 mō te c i te ture pūrua.

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

Mahia ngā tātaitai.

$$x=\frac{1±\sqrt{-7}}{4}$$

Tā te mea e kore te pūrua o tētahi tau tōraro e tautohutia ki te āpure tūturu, kāhore he rongoā.

$$x\in \emptyset$$

Rārangitia ngā otinga katoa i kitea.

$$x=-1$$