Whakaoti i te 3=1/x^2:x+4/2x:x

$$3 = \frac { 1 } { x ^ { 2 } } : x + \frac { 4 } { 2 x } : x$$

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.

$$3x=\frac{1}{x^{2}}+\frac{4}{2x}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o x^{2} me 2x ko 2x^{2}. Whakareatia \frac{1}{x^{2}} ki te \frac{2}{2}. Whakareatia \frac{4}{2x} ki te \frac{x}{x}.

$$3x=\frac{2}{2x^{2}}+\frac{4x}{2x^{2}}$$

Tā te mea he rite te tauraro o \frac{2}{2x^{2}} me \frac{4x}{2x^{2}}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$3x=\frac{2+4x}{2x^{2}}$$

Me whakatauwehe ngā kīanga kāore anō i whakatauwehea i roto o \frac{2+4x}{2x^{2}}.

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

Me whakakore tahi te 2 i te taurunga me te tauraro.

$$3x=\frac{2x+1}{x^{2}}$$

Tangohia te \frac{2x+1}{x^{2}} mai i ngā taha e rua.

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

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 3x ki te \frac{x^{2}}{x^{2}}.

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

Tā te mea he rite te tauraro o \frac{3xx^{2}}{x^{2}} me \frac{2x+1}{x^{2}}, me tango rāua mā te tango i ō raua taurunga.

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

Mahia ngā whakarea i roto o 3xx^{2}-\left(2x+1\right).

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

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}.

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

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.

$$3x=\frac{1}{x^{2}}+\frac{4}{2x}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o x^{2} me 2x ko 2x^{2}. Whakareatia \frac{1}{x^{2}} ki te \frac{2}{2}. Whakareatia \frac{4}{2x} ki te \frac{x}{x}.

$$3x=\frac{2}{2x^{2}}+\frac{4x}{2x^{2}}$$

Tā te mea he rite te tauraro o \frac{2}{2x^{2}} me \frac{4x}{2x^{2}}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$3x=\frac{2+4x}{2x^{2}}$$

Me whakatauwehe ngā kīanga kāore anō i whakatauwehea i roto o \frac{2+4x}{2x^{2}}.

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

Me whakakore tahi te 2 i te taurunga me te tauraro.

$$3x=\frac{2x+1}{x^{2}}$$

Tangohia te \frac{2x+1}{x^{2}} mai i ngā taha e rua.

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

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 3x ki te \frac{x^{2}}{x^{2}}.

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

Tā te mea he rite te tauraro o \frac{3xx^{2}}{x^{2}} me \frac{2x+1}{x^{2}}, me tango rāua mā te tango i ō raua taurunga.

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

Mahia ngā whakarea i roto o 3xx^{2}-\left(2x+1\right).

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

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}.

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

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.

$$x=1$$