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

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

Whakaoti mō x

$$x=-2$$

$$x=0$$

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), arā, te tauraro pātahi he tino iti rawa te kitea o 1+x,1-x.

$$-2\left(x-1\right)\left(x+1\right)=x-1-\left(-\left(1+x\right)\times 3\right)$$

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

$$\left(-2x+2\right)\left(x+1\right)=x-1-\left(-\left(1+x\right)\times 3\right)$$

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

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

Whakareatia te -1 ki te 3, ka -3.

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

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

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

Hei kimi i te tauaro o -3-3x, kimihia te tauaro o ia taurangi.

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

Tāpirihia te -1 ki te 3, ka 2.

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

Pahekotia te x me 3x, ka 4x.

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

Tangohia te 4x mai i ngā taha e rua.

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

Tangohia te 2 mai i ngā taha e rua.

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

Tangohia te 2 i te 2, ka 0.

$$-2x^{2}-4x=0$$

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

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

Tuhia te pūtakerua o te \left(-4\right)^{2}.

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

Ko te tauaro o -4 ko 4.

$$x=\frac{4±4}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{4±4}{-4}$$

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

$$x=\frac{8}{-4}$$

Whakawehe 8 ki te -4.

$$x=-2$$

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

$$x=\frac{0}{-4}$$

Whakawehe 0 ki te -4.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=-2 x=0$$

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), arā, te tauraro pātahi he tino iti rawa te kitea o 1+x,1-x.

$$-2\left(x-1\right)\left(x+1\right)=x-1-\left(-\left(1+x\right)\times 3\right)$$

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

$$\left(-2x+2\right)\left(x+1\right)=x-1-\left(-\left(1+x\right)\times 3\right)$$

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

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

Whakareatia te -1 ki te 3, ka -3.

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

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

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

Hei kimi i te tauaro o -3-3x, kimihia te tauaro o ia taurangi.

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

Tāpirihia te -1 ki te 3, ka 2.

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

Pahekotia te x me 3x, ka 4x.

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

Tangohia te 4x mai i ngā taha e rua.

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

Tangohia te 2 mai i ngā taha e rua.

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

Tangohia te 2 i te 2, ka 0.

$$-2x^{2}-4x=0$$

Whakawehea ngā taha e rua ki te -2.

$$\frac{-2x^{2}-4x}{-2}=\frac{0}{-2}$$

Mā te whakawehe ki te -2 ka wetekia te whakareanga ki te -2.

$$x^{2}+\left(-\frac{4}{-2}\right)x=\frac{0}{-2}$$

Whakawehe -4 ki te -2.

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

Whakawehe 0 ki te -2.

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

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.

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

Pūrua 1.

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

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}.

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

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

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

Whakarūnātia.

$$x+1=1 x+1=-1$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$x=0 x=-2$$