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

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

Whakaoti mō x

$$x = \frac{\sqrt{7} + 1}{2} \approx 1.822875656$$

$$x=\frac{1-\sqrt{7}}{2}\approx -0.822875656$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 2 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te -x+2.

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

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

$$x-1=-2x^{2}+4x-x+2$$

Pahekotia te 4x me -x, ka 3x.

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

Me tāpiri te 2x^{2} ki ngā taha e rua.

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

Tangohia te 3x mai i ngā taha e rua.

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

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

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

Tangohia te 2 mai i ngā taha e rua.

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

Tangohia te 2 i te -1, ka -3.

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

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.

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

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

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

Pūrua -2.

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

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-2\right)±\sqrt{4-8\left(-3\right)}}{2\times 2}$$

Whakareatia -8 ki te -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+24}}{2\times 2}$$

Tāpiri 4 ki te 24.

$$x=\frac{-\left(-2\right)±\sqrt{28}}{2\times 2}$$

Tuhia te pūtakerua o te 28.

$$x=\frac{-\left(-2\right)±2\sqrt{7}}{2\times 2}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2\sqrt{7}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{2±2\sqrt{7}}{4}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{7}}{4} ina he tāpiri te ±. Tāpiri 2 ki te 2\sqrt{7}.

$$x=\frac{2\sqrt{7}+2}{4}$$

Whakawehe 2+2\sqrt{7} ki te 4.

$$x=\frac{\sqrt{7}+1}{2}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{7}}{4} ina he tango te ±. Tango 2\sqrt{7} mai i 2.

$$x=\frac{2-2\sqrt{7}}{4}$$

Whakawehe 2-2\sqrt{7} ki te 4.

$$x=\frac{1-\sqrt{7}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{7}+1}{2} x=\frac{1-\sqrt{7}}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki 2 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te -x+2.

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

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

$$x-1=-2x^{2}+4x-x+2$$

Pahekotia te 4x me -x, ka 3x.

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

Me tāpiri te 2x^{2} ki ngā taha e rua.

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

Tangohia te 3x mai i ngā taha e rua.

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

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

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

Me tāpiri te 1 ki ngā taha e rua.

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

Tāpirihia te 2 ki te 1, ka 3.

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

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.

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

Whakawehea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

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

Whakawehe -2 ki te 2.

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

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.

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

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

$$x^{2}-x+\frac{1}{4}=\frac{3}{2}+\frac{1}{4}$$

Tāpiri \frac{3}{2} ki te \frac{1}{4} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}-x+\frac{1}{4}=\frac{7}{4}$$

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

$$\left(x-\frac{1}{2}\right)^{2}=\frac{7}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{7}{4}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{\sqrt{7}}{2} x-\frac{1}{2}=-\frac{\sqrt{7}}{2}$$

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

$$x=\frac{\sqrt{7}+1}{2} x=\frac{1-\sqrt{7}}{2}$$