Whakaoti i te 1/x+1+2/x-1=x^2+2x/x^2-1

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

Whakaoti mō x

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

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-1\right)\left(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+1,x-1,x^{2}-1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 2.

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

Pahekotia te x me 2x, ka 3x.

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

Tāpirihia te -1 ki te 2, ka 1.

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

Tangohia te x^{2} mai i ngā taha e rua.

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

Tangohia te 2x mai i ngā taha e rua.

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

Pahekotia te 3x me -2x, ka x.

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

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

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

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}.

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

Pūrua 1.

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

Whakareatia -4 ki te -1.

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

Tāpiri 1 ki te 4.

$$x=\frac{-1±\sqrt{5}}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-1±\sqrt{5}}{-2}$$

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

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

Whakawehe -1+\sqrt{5} ki te -2.

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

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

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

Whakawehe -1-\sqrt{5} ki te -2.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-1\right)\left(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+1,x-1,x^{2}-1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 2.

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

Pahekotia te x me 2x, ka 3x.

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

Tāpirihia te -1 ki te 2, ka 1.

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

Tangohia te x^{2} mai i ngā taha e rua.

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

Tangohia te 2x mai i ngā taha e rua.

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

Pahekotia te 3x me -2x, ka x.

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

Tangohia te 1 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

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

Whakawehea ngā taha e rua ki te -1.

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

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

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

Whakawehe 1 ki te -1.

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

Whakawehe -1 ki te -1.

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

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 tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

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