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

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

Whakaoti mō x

$$x=-2$$

Ngā Upane Whakamahi Tauwehe

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

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te 2x-1 ka whakakotahi i ngā kupu rite.

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

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

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

Whakaarohia te \left(x-1\right)\left(x+1\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 1.

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

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

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

Pahekotia te 2x^{2} me -x^{2}, ka x^{2}.

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

Me tāpiri te 1 ki ngā taha e rua.

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

Tāpirihia te -3 ki te 1, ka -2.

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

Hei whakaoti i te whārite, whakatauwehea te x^{2}+x-2 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.

$$a+b=1 ab=-2$$

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

$$a=-1 b=2$$

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

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

Hei kimi otinga whārite, me whakaoti te x-1=0 me te x+2=0.

$$x=1 x=-2$$

Tē taea kia ōrite te tāupe x ki 1.

$$x=-2$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te 2x-1 ka whakakotahi i ngā kupu rite.

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

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

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

Whakaarohia te \left(x-1\right)\left(x+1\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 1.

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

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

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

Pahekotia te 2x^{2} me -x^{2}, ka x^{2}.

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

Me tāpiri te 1 ki ngā taha e rua.

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

Tāpirihia te -3 ki te 1, ka -2.

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

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

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

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

$$a=-1 b=2$$

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

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

Tauwehea te x i te tuatahi me te 2 i te rōpū tuarua.

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

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

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

Hei kimi otinga whārite, me whakaoti te x-1=0 me te x+2=0.

$$x=1 x=-2$$

Tē taea kia ōrite te tāupe x ki 1.

$$x=-2$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te 2x-1 ka whakakotahi i ngā kupu rite.

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

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

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

Whakaarohia te \left(x-1\right)\left(x+1\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 1.

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

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

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

Pahekotia te 2x^{2} me -x^{2}, ka x^{2}.

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

Me tāpiri te 1 ki ngā taha e rua.

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

Tāpirihia te -3 ki te 1, ka -2.

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

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

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

Pūrua 1.

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

Whakareatia -4 ki te -2.

$$x=\frac{-1±\sqrt{1+8}}{2}$$

Tāpiri 1 ki te 8.

$$x=\frac{-1±\sqrt{9}}{2}$$

Tuhia te pūtakerua o te 9.

$$x=\frac{-1±3}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±3}{2} ina he tāpiri te ±. Tāpiri -1 ki te 3.

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

Whakawehe 2 ki te 2.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{-1±3}{2} ina he tango te ±. Tango 3 mai i -1.

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

Whakawehe -4 ki te 2.

$$x=-2$$

Kua oti te whārite te whakatau.

$$x=1 x=-2$$

Tē taea kia ōrite te tāupe x ki 1.

$$x=-2$$

Ngā Upane mō te Whakaoti i te Pūrua

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

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te 2x-1 ka whakakotahi i ngā kupu rite.

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

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

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

Whakaarohia te \left(x-1\right)\left(x+1\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 1.

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

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

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

Pahekotia te 2x^{2} me -x^{2}, ka x^{2}.

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

Me tāpiri te 3 ki ngā taha e rua.

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

Tāpirihia te -1 ki te 3, ka 2.

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

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.

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

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

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

Tāpiri 2 ki te \frac{1}{4}.

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

Tauwehea x^{2}+x+\frac{1}{4}. 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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{9}{4}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Whakarūnātia.

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

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$x=1 x=-2$$

Tē taea kia ōrite te tāupe x ki 1.

$$x=-2$$