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

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

Whakaoti mō x

$$x=0$$

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)^{2}, arā, te tauraro pātahi he tino iti rawa te kitea o x^{3}-x^{2}-x+1,1-x^{2},x+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 1-x ki te 3.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-1\right)^{2}.

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

Hei kimi i te tauaro o x^{2}-2x+1, kimihia te tauaro o ia taurangi.

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

Pahekotia te -3x me 2x, ka -x.

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

Tangohia te 1 i te 3, ka 2.

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

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

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

Tangohia te 2 mai i ngā taha e rua.

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

Tangohia te 2 i te 2, ka 0.

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

Tauwehea te x.

$$x\left(-1-x\right)=0$$

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

$$x=0 x=-1$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=0$$

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)^{2}, arā, te tauraro pātahi he tino iti rawa te kitea o x^{3}-x^{2}-x+1,1-x^{2},x+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 1-x ki te 3.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-1\right)^{2}.

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

Hei kimi i te tauaro o x^{2}-2x+1, kimihia te tauaro o ia taurangi.

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

Pahekotia te -3x me 2x, ka -x.

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

Tangohia te 1 i te 3, ka 2.

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

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

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

Tangohia te 2 mai i ngā taha e rua.

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

Tangohia te 2 i te 2, ka 0.

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

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.

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

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

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

Tuhia te pūtakerua o te 1.

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

Ko te tauaro o -1 ko 1.

$$x=\frac{1±1}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{1±1}{-2}$$

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

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

Whakawehe 2 ki te -2.

$$x=-1$$

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

$$x=\frac{0}{-2}$$

Whakawehe 0 ki te -2.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=-1 x=0$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=0$$

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)^{2}, arā, te tauraro pātahi he tino iti rawa te kitea o x^{3}-x^{2}-x+1,1-x^{2},x+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 1-x ki te 3.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-1\right)^{2}.

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

Hei kimi i te tauaro o x^{2}-2x+1, kimihia te tauaro o ia taurangi.

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

Pahekotia te -3x me 2x, ka -x.

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

Tangohia te 1 i te 3, ka 2.

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

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

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

Tangohia te 2 mai i ngā taha e rua.

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

Tangohia te 2 i te 2, ka 0.

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

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.

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

Whakawehea ngā taha e rua ki te -1.

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

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

$$x^{2}+\left(-\frac{1}{-1}\right)x=\frac{0}{-1}$$

Whakawehe -1 ki te -1.

$$x^{2}+x=\frac{0}{-1}$$

Whakawehe 0 ki te -1.

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

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.

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

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

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

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}.

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

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

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

Whakarūnātia.

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

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

$$x=0 x=-1$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=0$$