Whakaoti i te 3x^2-1/x-2=0

$$3 x ^ { 2 } - \frac { 1 } { x } - 2 = 0$$

Whakaoti mō x (complex solution)

$$x=-\frac{\sqrt{3}i}{6}-\frac{1}{2}\approx -0.5-0.288675135i$$

$$x=1$$

$$x=\frac{\sqrt{3}i}{6}-\frac{1}{2}\approx -0.5+0.288675135i$$

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.

$$3x^{2}x-1+x\left(-2\right)=0$$

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

$$3x^{3}-1+x\left(-2\right)=0$$

Hurinahatia te whārite ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

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

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 3. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{1}{3},±1$$

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 3x^{3}-2x-1 ki te x-1, kia riro ko 3x^{2}+3x+1. Whakaotihia te whārite ina ōrite te hua ki te 0.

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

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 3 mō te a, te 3 mō te b, me te 1 mō te c i te ture pūrua.

$$x=\frac{-3±\sqrt{3^{2}-4\times 3\times 1}}{2\times 3}$$

Mahia ngā tātaitai.

$$x=\frac{-3±\sqrt{-3}}{6}$$

Whakaotia te whārite 3x^{2}+3x+1=0 ina he tōrunga te ±, ina he tōraro te ±.

$$x=-\frac{\sqrt{3}i}{6}-\frac{1}{2} x=\frac{\sqrt{3}i}{6}-\frac{1}{2}$$

Rārangitia ngā otinga katoa i kitea.

$$x=1 x=-\frac{\sqrt{3}i}{6}-\frac{1}{2} x=\frac{\sqrt{3}i}{6}-\frac{1}{2}$$

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.

$$3x^{2}x-1+x\left(-2\right)=0$$

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

$$3x^{3}-1+x\left(-2\right)=0$$

Hurinahatia te whārite ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

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

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 3. Whakarārangitia ngā kaitono katoa \frac{p}{q}.

$$±\frac{1}{3},±1$$

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 3x^{3}-2x-1 ki te x-1, kia riro ko 3x^{2}+3x+1. Whakaotihia te whārite ina ōrite te hua ki te 0.

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

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 3 mō te a, te 3 mō te b, me te 1 mō te c i te ture pūrua.

$$x=\frac{-3±\sqrt{3^{2}-4\times 3\times 1}}{2\times 3}$$

Mahia ngā tātaitai.

$$x=\frac{-3±\sqrt{-3}}{6}$$

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.

$$x=1$$