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

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

Whakaoti mō x

$$x=\frac{\sqrt{3}}{3}+1\approx 1.577350269$$

$$x=-\frac{\sqrt{3}}{3}+1\approx 0.422649731$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara 0,2 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(x-2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x,x-2.

$$x-2-x=3x\left(x-2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te x-2.

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

Tangohia te 3x^{2} mai i ngā taha e rua.

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

Me tāpiri te 6x ki ngā taha e rua.

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

Pahekotia te x me 6x, ka 7x.

$$7x-2-x-3x^{2}=0$$

Pahekotia te 7x me -x, ka 6x.

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

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.

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

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

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

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\left(-3\right)\left(-2\right)}}{2\left(-3\right)}$$

Whakareatia -4 ki te -3.

$$x=\frac{-6±\sqrt{36+12\left(-2\right)}}{2\left(-3\right)}$$

Whakareatia 12 ki te -2.

$$x=\frac{-6±\sqrt{36-24}}{2\left(-3\right)}$$

Tāpiri 36 ki te -24.

$$x=\frac{-6±\sqrt{12}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te 12.

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

Whakareatia 2 ki te -3.

$$x=\frac{-6±2\sqrt{3}}{-6}$$

Nā, me whakaoti te whārite x=\frac{-6±2\sqrt{3}}{-6} ina he tāpiri te ±. Tāpiri -6 ki te 2\sqrt{3}.

$$x=\frac{2\sqrt{3}-6}{-6}$$

Whakawehe -6+2\sqrt{3} ki te -6.

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

Nā, me whakaoti te whārite x=\frac{-6±2\sqrt{3}}{-6} ina he tango te ±. Tango 2\sqrt{3} mai i -6.

$$x=\frac{-2\sqrt{3}-6}{-6}$$

Whakawehe -6-2\sqrt{3} ki te -6.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara 0,2 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(x-2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x,x-2.

$$x-2-x=3x\left(x-2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te x-2.

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

Tangohia te 3x^{2} mai i ngā taha e rua.

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

Me tāpiri te 6x ki ngā taha e rua.

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

Pahekotia te x me 6x, ka 7x.

$$7x-2-x-3x^{2}=0$$

Me tāpiri te 2 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$7x-x-3x^{2}=2$$

Pahekotia te 7x me -x, ka 6x.

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

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.

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

Whakawehea ngā taha e rua ki te -3.

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

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

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

Whakawehe 6 ki te -3.

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

Whakawehe 2 ki te -3.

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

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.

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

Tāpiri -\frac{2}{3} ki te 1.

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

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}.

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

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

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

Whakarūnātia.

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

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

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