Whakaoti i te (x)=x-2/x-1

$$( x ) = \frac { x - 2 } { x - 1 }$$

Whakaoti mō x (complex solution)

$$x=1+i$$

$$x=1-i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

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

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

$$\frac{x\left(x-1\right)}{x-1}-\frac{x-2}{x-1}=0$$

Tā te mea he rite te tauraro o \frac{x\left(x-1\right)}{x-1} me \frac{x-2}{x-1}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{x\left(x-1\right)-\left(x-2\right)}{x-1}=0$$

Mahia ngā whakarea i roto o x\left(x-1\right)-\left(x-2\right).

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

Whakakotahitia ngā kupu rite i x^{2}-x-x+2.

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

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.

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

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

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

Pūrua -2.

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

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-2\right)±\sqrt{4-8}}{2}$$

Tāpiri 4 ki te -8.

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

Tuhia te pūtakerua o te -4.

$$x=\frac{-\left(-2\right)±2i}{2}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2i}{2}$$

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

$$x=\frac{2+2i}{2}$$

Whakawehe 2+2i ki te 2.

$$x=1+i$$

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

$$x=\frac{2-2i}{2}$$

Whakawehe 2-2i ki te 2.

$$x=1-i$$

Kua oti te whārite te whakatau.

$$x=1+i x=1-i$$

Ngā Upane mō te Whakaoti i te Pūrua

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

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

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

$$\frac{x\left(x-1\right)}{x-1}-\frac{x-2}{x-1}=0$$

Tā te mea he rite te tauraro o \frac{x\left(x-1\right)}{x-1} me \frac{x-2}{x-1}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{x\left(x-1\right)-\left(x-2\right)}{x-1}=0$$

Mahia ngā whakarea i roto o x\left(x-1\right)-\left(x-2\right).

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

Whakakotahitia ngā kupu rite i x^{2}-x-x+2.

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

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.

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

Tangohia te 2 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$x^{2}-2x=-2$$

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

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

Tāpiri -2 ki te 1.

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

Tauwehea x^{2}-2x+1. 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}.

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

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

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

Whakarūnātia.

$$x-1=i x-1=-i$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$x=1+i x=1-i$$