Whakaoti i te 3x/x-1-4/x=3/x^2-x

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

Whakaoti mō x

$$x=\frac{1}{3}\approx 0.333333333$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara 0,1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(x-1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x-1,x,x^{2}-x.

$$x\times 3x-\left(x-1\right)\times 4=3$$

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

$$x^{2}\times 3-\left(x-1\right)\times 4=3$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-1 ki te 4.

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

Hei kimi i te tauaro o 4x-4, kimihia te tauaro o ia taurangi.

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

Tangohia te 3 mai i ngā taha e rua.

$$x^{2}\times 3-4x+4-3=0$$

Tangohia te 3 i te 4, ka 1.

$$x^{2}\times 3-4x+1=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 3x^{2}+ax+bx+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=3\times 1=3$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-3 b=-1$$

Tuhia anō te 3x^{2}-4x+1 hei \left(3x^{2}-3x\right)+\left(-x+1\right).

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

Tauwehea te 3x i te tuatahi me te -1 i te rōpū tuarua.

$$3x\left(x-1\right)-\left(x-1\right)$$

Whakatauwehea atu te kīanga pātahi x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-1\right)\left(3x-1\right)$$

Hei kimi otinga whārite, me whakaoti te x-1=0 me te 3x-1=0.

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

Tē taea kia ōrite te tāupe x ki 1.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara 0,1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(x-1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x-1,x,x^{2}-x.

$$x\times 3x-\left(x-1\right)\times 4=3$$

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

$$x^{2}\times 3-\left(x-1\right)\times 4=3$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-1 ki te 4.

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

Hei kimi i te tauaro o 4x-4, kimihia te tauaro o ia taurangi.

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

Tangohia te 3 mai i ngā taha e rua.

$$x^{2}\times 3-4x+4-3=0$$

Tangohia te 3 i te 4, ka 1.

$$x^{2}\times 3-4x+1=0$$

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.

$$3x^{2}-4x+1=0$$

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

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

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 3}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-4\right)±\sqrt{16-12}}{2\times 3}$$

Tāpiri 16 ki te -12.

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

Tuhia te pūtakerua o te 4.

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

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{4±2}{6}$$

Nā, me whakaoti te whārite x=\frac{4±2}{6} ina he tāpiri te ±. Tāpiri 4 ki te 2.

$$x=\frac{6}{6}$$

Whakawehe 6 ki te 6.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{4±2}{6} ina he tango te ±. Tango 2 mai i 4.

$$x=\frac{2}{6}$$

Whakahekea te hautanga \frac{2}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Kua oti te whārite te whakatau.

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

Tē taea kia ōrite te tāupe x ki 1.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara 0,1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(x-1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x-1,x,x^{2}-x.

$$x\times 3x-\left(x-1\right)\times 4=3$$

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

$$x^{2}\times 3-\left(x-1\right)\times 4=3$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-1 ki te 4.

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

Hei kimi i te tauaro o 4x-4, kimihia te tauaro o ia taurangi.

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

Tangohia te 4 mai i ngā taha e rua.

$$x^{2}\times 3-4x=3-4$$

Tangohia te 4 i te 3, ka -1.

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

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.

$$3x^{2}-4x=-1$$

Whakawehea ngā taha e rua ki te 3.

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

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

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

Whakawehea te -\frac{4}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{2}{3}. Nā, tāpiria te pūrua o te -\frac{2}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=-\frac{1}{3}+\left(-\frac{2}{3}\right)^{2}$$

Pūruatia -\frac{2}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=-\frac{1}{3}+\frac{4}{9}$$

Tāpiri -\frac{1}{3} ki te \frac{4}{9} 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.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{1}{9}$$

Tauwehea x^{2}-\frac{4}{3}x+\frac{4}{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}.

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

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

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

Whakarūnātia.

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

Me tāpiri \frac{2}{3} ki ngā taha e rua o te whārite.

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

Tē taea kia ōrite te tāupe x ki 1.

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