Whakaoti i te 1/x-1-x/x+1-2=0

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

Whakaoti mō x

$$x = \frac{\sqrt{10} + 1}{3} \approx 1.387425887$$

$$x=\frac{1-\sqrt{10}}{3}\approx -0.72075922$$

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.

$$x+1-\left(x-1\right)x+\left(x-1\right)\left(x+1\right)\left(-2\right)=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-1 ki te x.

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

Hei kimi i te tauaro o x^{2}-x, kimihia te tauaro o ia taurangi.

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

Pahekotia te x me x, ka 2x.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x-1 ki te x+1 ka whakakotahi i ngā kupu rite.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}-1 ki te -2.

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

Pahekotia te -x^{2} me -2x^{2}, ka -3x^{2}.

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

Tāpirihia te 1 ki te 2, ka 3.

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

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -3 mō a, 2 mō b, me 3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-3\right)\times 3}}{2\left(-3\right)}$$

Pūrua 2.

$$x=\frac{-2±\sqrt{4-4\left(-3\right)\times 3}}{2\left(-3\right)}$$

Whakareatia -4 ki te -3.

$$x=\frac{-2±\sqrt{4+12\times 3}}{2\left(-3\right)}$$

Whakareatia 12 ki te 3.

$$x=\frac{-2±\sqrt{4+36}}{2\left(-3\right)}$$

Tāpiri 4 ki te 36.

$$x=\frac{-2±\sqrt{40}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te 40.

$$x=\frac{-2±2\sqrt{10}}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{-2±2\sqrt{10}}{-6}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{10}}{-6} ina he tāpiri te ±. Tāpiri -2 ki te 2\sqrt{10}.

$$x=\frac{2\sqrt{10}-2}{-6}$$

Whakawehe -2+2\sqrt{10} ki te -6.

$$x=\frac{1-\sqrt{10}}{3}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{10}}{-6} ina he tango te ±. Tango 2\sqrt{10} mai i -2.

$$x=\frac{-2\sqrt{10}-2}{-6}$$

Whakawehe -2-2\sqrt{10} ki te -6.

$$x=\frac{\sqrt{10}+1}{3}$$

Kua oti te whārite te whakatau.

$$x=\frac{1-\sqrt{10}}{3} x=\frac{\sqrt{10}+1}{3}$$

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.

$$x+1-\left(x-1\right)x+\left(x-1\right)\left(x+1\right)\left(-2\right)=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-1 ki te x.

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

Hei kimi i te tauaro o x^{2}-x, kimihia te tauaro o ia taurangi.

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

Pahekotia te x me x, ka 2x.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x-1 ki te x+1 ka whakakotahi i ngā kupu rite.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}-1 ki te -2.

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

Pahekotia te -x^{2} me -2x^{2}, ka -3x^{2}.

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

Tāpirihia te 1 ki te 2, ka 3.

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

Tangohia te 3 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

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.

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

Whakawehea ngā taha e rua ki te -3.

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

Mā te whakawehe ki te -3 ka wetekia te whakareanga ki te -3.

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

Whakawehe 2 ki te -3.

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

Whakawehe -3 ki te -3.

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

Whakawehea te -\frac{2}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{3}. Nā, tāpiria te pūrua o te -\frac{1}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{1}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=1+\frac{1}{9}$$

Tāpiri 1 ki te \frac{1}{9}.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{10}{9}$$

Tauwehea x^{2}-\frac{2}{3}x+\frac{1}{9}. 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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{10}{9}$$

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

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{10}{9}}$$

Whakarūnātia.

$$x-\frac{1}{3}=\frac{\sqrt{10}}{3} x-\frac{1}{3}=-\frac{\sqrt{10}}{3}$$

Me tāpiri \frac{1}{3} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{10}+1}{3} x=\frac{1-\sqrt{10}}{3}$$