Whakaoti i te x=1/x+1

$$x= \frac{ 1 }{ x+1 }$$

Whakaoti mō x

$$x=\frac{\sqrt{5}-1}{2}\approx 0.618033989$$

$$x=\frac{-\sqrt{5}-1}{2}\approx -1.618033989$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$x-\frac{1}{x+1}=0$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia x ki te \frac{x+1}{x+1}.

$$\frac{x\left(x+1\right)}{x+1}-\frac{1}{x+1}=0$$

Tā te mea he rite te tauraro o \frac{x\left(x+1\right)}{x+1} me \frac{1}{x+1}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{x\left(x+1\right)-1}{x+1}=0$$

Mahia ngā whakarea i roto o x\left(x+1\right)-1.

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

Tē taea kia ōrite te tāupe x ki -1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x+1.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 1 mō b, me -1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua 1.

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

Whakareatia -4 ki te -1.

$$x=\frac{-1±\sqrt{1+4}}{2}$$

Tāpiri 1 ki te 4.

$$x=\frac{-1±\sqrt{5}}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{5}}{2} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{5}.

$$x=\frac{\sqrt{5}-1}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{5}}{2} ina he tango te ±. Tango \sqrt{5} mai i -1.

$$x=\frac{-\sqrt{5}-1}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{5}-1}{2} x=\frac{-\sqrt{5}-1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

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

$$x-\frac{1}{x+1}=0$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia x ki te \frac{x+1}{x+1}.

$$\frac{x\left(x+1\right)}{x+1}-\frac{1}{x+1}=0$$

Tā te mea he rite te tauraro o \frac{x\left(x+1\right)}{x+1} me \frac{1}{x+1}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{x\left(x+1\right)-1}{x+1}=0$$

Mahia ngā whakarea i roto o x\left(x+1\right)-1.

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

Tē taea kia ōrite te tāupe x ki -1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x+1.

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

Me tāpiri te 1 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

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

Whakawehea te 1, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{2}. Nā, tāpiria te pūrua o te \frac{1}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri 1 ki te \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{5}{4}$$

Tauwehea x^{2}+x+\frac{1}{4}. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{5}{4}$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Whakarūnātia.

$$x+\frac{1}{2}=\frac{\sqrt{5}}{2} x+\frac{1}{2}=-\frac{\sqrt{5}}{2}$$

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{5}-1}{2} x=\frac{-\sqrt{5}-1}{2}$$