Whakaoti i te 1/3x-2=2x-3

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

Whakaoti mō x

$$x=\frac{\sqrt{33}}{12}+\frac{1}{4}\approx 0.728713554$$

$$x=-\frac{\sqrt{33}}{12}+\frac{1}{4}\approx -0.228713554$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te 3x.

$$1+3x\left(-2\right)=2x\times 3x+3x\left(-3\right)$$

Whakareatia te 3 ki te -2, ka -6.

$$1-6x=2x\times 3x+3x\left(-3\right)$$

Whakareatia te x ki te x, ka x^{2}.

$$1-6x=2x^{2}\times 3+3x\left(-3\right)$$

Whakareatia te 2 ki te 3, ka 6.

$$1-6x=6x^{2}+3x\left(-3\right)$$

Whakareatia te 3 ki te -3, ka -9.

$$1-6x=6x^{2}-9x$$

Tangohia te 6x^{2} mai i ngā taha e rua.

$$1-6x-6x^{2}=-9x$$

Me tāpiri te 9x ki ngā taha e rua.

$$1-6x-6x^{2}+9x=0$$

Pahekotia te -6x me 9x, ka 3x.

$$1+3x-6x^{2}=0$$

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.

$$-6x^{2}+3x+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -6 mō a, 3 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-6\right)}}{2\left(-6\right)}$$

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\left(-6\right)}}{2\left(-6\right)}$$

Whakareatia -4 ki te -6.

$$x=\frac{-3±\sqrt{9+24}}{2\left(-6\right)}$$

Tāpiri 9 ki te 24.

$$x=\frac{-3±\sqrt{33}}{2\left(-6\right)}$$

Whakareatia 2 ki te -6.

$$x=\frac{-3±\sqrt{33}}{-12}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{33}}{-12} ina he tāpiri te ±. Tāpiri -3 ki te \sqrt{33}.

$$x=\frac{\sqrt{33}-3}{-12}$$

Whakawehe -3+\sqrt{33} ki te -12.

$$x=-\frac{\sqrt{33}}{12}+\frac{1}{4}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{33}}{-12} ina he tango te ±. Tango \sqrt{33} mai i -3.

$$x=\frac{-\sqrt{33}-3}{-12}$$

Whakawehe -3-\sqrt{33} ki te -12.

$$x=\frac{\sqrt{33}}{12}+\frac{1}{4}$$

Kua oti te whārite te whakatau.

$$x=-\frac{\sqrt{33}}{12}+\frac{1}{4} x=\frac{\sqrt{33}}{12}+\frac{1}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te 3x.

$$1+3x\left(-2\right)=2x\times 3x+3x\left(-3\right)$$

Whakareatia te 3 ki te -2, ka -6.

$$1-6x=2x\times 3x+3x\left(-3\right)$$

Whakareatia te x ki te x, ka x^{2}.

$$1-6x=2x^{2}\times 3+3x\left(-3\right)$$

Whakareatia te 2 ki te 3, ka 6.

$$1-6x=6x^{2}+3x\left(-3\right)$$

Whakareatia te 3 ki te -3, ka -9.

$$1-6x=6x^{2}-9x$$

Tangohia te 6x^{2} mai i ngā taha e rua.

$$1-6x-6x^{2}=-9x$$

Me tāpiri te 9x ki ngā taha e rua.

$$1-6x-6x^{2}+9x=0$$

Pahekotia te -6x me 9x, ka 3x.

$$1+3x-6x^{2}=0$$

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

$$3x-6x^{2}=-1$$

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.

$$-6x^{2}+3x=-1$$

Whakawehea ngā taha e rua ki te -6.

$$\frac{-6x^{2}+3x}{-6}=-\frac{1}{-6}$$

Mā te whakawehe ki te -6 ka wetekia te whakareanga ki te -6.

$$x^{2}+\frac{3}{-6}x=-\frac{1}{-6}$$

Whakahekea te hautanga \frac{3}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$x^{2}-\frac{1}{2}x=-\frac{1}{-6}$$

Whakawehe -1 ki te -6.

$$x^{2}-\frac{1}{2}x=\frac{1}{6}$$

Whakawehea te -\frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{4}. Nā, tāpiria te pūrua o te -\frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{1}{6}+\left(-\frac{1}{4}\right)^{2}$$

Pūruatia -\frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{6}+\frac{1}{16}$$

Tāpiri \frac{1}{6} ki te \frac{1}{16} 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.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{11}{48}$$

Tauwehea x^{2}-\frac{1}{2}x+\frac{1}{16}. 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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{11}{48}$$

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

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{11}{48}}$$

Whakarūnātia.

$$x-\frac{1}{4}=\frac{\sqrt{33}}{12} x-\frac{1}{4}=-\frac{\sqrt{33}}{12}$$

Me tāpiri \frac{1}{4} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{33}}{12}+\frac{1}{4} x=-\frac{\sqrt{33}}{12}+\frac{1}{4}$$