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

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

Whakaoti mō x

$$x=3$$

Ngā Upane Whakamahi Tauwehe

Tē taea kia ōrite te tāupe x ki -1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x+1.

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

Tangohia te 2x mai i ngā taha e rua.

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

Tangohia te 2 mai i ngā taha e rua.

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

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

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

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

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

Hei whakaoti i te whārite, whakatauwehea te x^{2}-2x-3 mā te whakamahi i te tātai x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-2 ab=-3$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-3 b=1$$

Me tuhi anō te kīanga whakatauwehe \left(x+a\right)\left(x+b\right) mā ngā uara i tātaihia.

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

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

$$x=3 x=-1$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=3$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe x ki -1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x+1.

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

Tangohia te 2x mai i ngā taha e rua.

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

Tangohia te 2 mai i ngā taha e rua.

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

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

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

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

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

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

$$a+b=-2 ab=1\left(-3\right)=-3$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-3 b=1$$

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

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

Whakatauwehea atu x i te x^{2}-3x.

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

Whakatauwehea atu te kīanga pātahi x-3 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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

$$x=3 x=-1$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=3$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki -1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x+1.

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

Tangohia te 2x mai i ngā taha e rua.

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

Tangohia te 2 mai i ngā taha e rua.

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

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

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

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.

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

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

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

Pūrua -2.

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

Whakareatia -4 ki te -3.

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

Tāpiri 4 ki te 12.

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

Tuhia te pūtakerua o te 16.

$$x=\frac{-\left(-2\right)±4}{2}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±4}{2}$$

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

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

Whakawehe 6 ki te 2.

$$x=3$$

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

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

Whakawehe -2 ki te 2.

$$x=-1$$

Kua oti te whārite te whakatau.

$$x=3 x=-1$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=3$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki -1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x+1.

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

Tangohia te 2x mai i ngā taha e rua.

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

Me tāpiri te 1 ki ngā taha e rua.

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

Tāpirihia te 2 ki te 1, ka 3.

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

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.

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

Tāpiri 3 ki te 1.

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

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}.

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

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

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

Whakarūnātia.

$$x-1=2 x-1=-2$$

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

$$x=3 x=-1$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=3$$