Whakaoti i te x=sqrt{x}*1/x

$$x = \sqrt { x } \times \frac { 1 } { x }$$

Whakaoti mō x (complex solution)

$$x=\frac{-\sqrt{3}i-1}{2}\approx -0.5-0.866025404i$$

$$x=1$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Tuhia te \sqrt{x}\times \frac{1}{x} hei hautanga kotahi.

$$x^{2}=\left(\frac{\sqrt{x}}{x}\right)^{2}$$

Kia whakarewa i te \frac{\sqrt{x}}{x} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$x^{2}=\frac{\left(\sqrt{x}\right)^{2}}{x^{2}}$$

Tātaihia te \sqrt{x} mā te pū o 2, kia riro ko x.

$$x^{2}=\frac{x}{x^{2}}$$

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

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

Whakareatia ngā taha e rua o te whārite ki te x.

$$xx^{2}=1$$

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.

$$x^{3}=1$$

Tangohia te 1 mai i ngā taha e rua.

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

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

$$±1$$

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

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

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

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

Mahia ngā tātaitai.

$$x=\frac{-1±\sqrt{-3}}{2}$$

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

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

Rārangitia ngā otinga katoa i kitea.

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

Whakakapia te 1 mō te x i te whārite x=\sqrt{x}\times \frac{1}{x}.

$$1=\sqrt{1}\times \frac{1}{1}$$

Whakarūnātia. Ko te uara x=1 kua ngata te whārite.

$$1=1$$

Whakakapia te \frac{-\sqrt{3}i-1}{2} mō te x i te whārite x=\sqrt{x}\times \frac{1}{x}.

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

Whakarūnātia. Ko te uara x=\frac{-\sqrt{3}i-1}{2} kua ngata te whārite.

$$-\frac{1}{2}i\times 3^{\frac{1}{2}}-\frac{1}{2}=-\frac{1}{2}-\frac{1}{2}i\times 3^{\frac{1}{2}}$$

Whakakapia te \frac{-1+\sqrt{3}i}{2} mō te x i te whārite x=\sqrt{x}\times \frac{1}{x}.

$$\frac{-1+\sqrt{3}i}{2}=\sqrt{\frac{-1+\sqrt{3}i}{2}}\times \frac{1}{\frac{-1+\sqrt{3}i}{2}}$$

Whakarūnātia. Ko te uara x=\frac{-1+\sqrt{3}i}{2} kāore e ngata ana ki te whārite.

$$-\frac{1}{2}+\frac{1}{2}i\times 3^{\frac{1}{2}}=\frac{1}{2}-\frac{1}{2}i\times 3^{\frac{1}{2}}$$

Rārangihia ngā rongoā katoa o x=\frac{1}{x}\sqrt{x}.

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

Whakaoti mō x

$$x=1$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Tuhia te \sqrt{x}\times \frac{1}{x} hei hautanga kotahi.

$$x^{2}=\left(\frac{\sqrt{x}}{x}\right)^{2}$$

Kia whakarewa i te \frac{\sqrt{x}}{x} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$x^{2}=\frac{\left(\sqrt{x}\right)^{2}}{x^{2}}$$

Tātaihia te \sqrt{x} mā te pū o 2, kia riro ko x.

$$x^{2}=\frac{x}{x^{2}}$$

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

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

Whakareatia ngā taha e rua o te whārite ki te x.

$$xx^{2}=1$$

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.

$$x^{3}=1$$

Tangohia te 1 mai i ngā taha e rua.

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

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

$$±1$$

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

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

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

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

Mahia ngā tātaitai.

$$x=\frac{-1±\sqrt{-3}}{2}$$

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.

$$x=1$$

Whakakapia te 1 mō te x i te whārite x=\sqrt{x}\times \frac{1}{x}.

$$1=\sqrt{1}\times \frac{1}{1}$$

Whakarūnātia. Ko te uara x=1 kua ngata te whārite.

$$1=1$$

Ko te whārite x=\frac{1}{x}\sqrt{x} he rongoā ahurei.

$$x=1$$