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

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

Whakaoti mō x

$$x = \frac{3 \sqrt{2}}{2} \approx 2.121320344$$

$$x = -\frac{3 \sqrt{2}}{2} \approx -2.121320344$$

Ngā Upane mā te Kimi Pūtakerua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,-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)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{3}+2x^{2}-x-2,1-x,x+1.

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

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

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

Tāpirihia te 1 ki te 2, ka 3.

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

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}+x-2 ki te 3.

$$3+3x+x^{2}=3x^{2}+3x-6$$

Tangohia te 3x^{2} mai i ngā taha e rua.

$$3+3x+x^{2}-3x^{2}=3x-6$$

Pahekotia te x^{2} me -3x^{2}, ka -2x^{2}.

$$3+3x-2x^{2}=3x-6$$

Tangohia te 3x mai i ngā taha e rua.

$$3+3x-2x^{2}-3x=-6$$

Pahekotia te 3x me -3x, ka 0.

$$3-2x^{2}=-6$$

Tangohia te 3 mai i ngā taha e rua.

$$-2x^{2}=-6-3$$

Tangohia te 3 i te -6, ka -9.

$$-2x^{2}=-9$$

Whakawehea ngā taha e rua ki te -2.

$$x^{2}=\frac{-9}{-2}$$

Ka taea te hautanga \frac{-9}{-2} te whakamāmā ki te \frac{9}{2} mā te tango tahi i te tohu tōraro i te taurunga me te tauraro.

$$x^{2}=\frac{9}{2}$$

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

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,-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)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{3}+2x^{2}-x-2,1-x,x+1.

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

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

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

Tāpirihia te 1 ki te 2, ka 3.

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

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}+x-2 ki te 3.

$$3+3x+x^{2}=3x^{2}+3x-6$$

Tangohia te 3x^{2} mai i ngā taha e rua.

$$3+3x+x^{2}-3x^{2}=3x-6$$

Pahekotia te x^{2} me -3x^{2}, ka -2x^{2}.

$$3+3x-2x^{2}=3x-6$$

Tangohia te 3x mai i ngā taha e rua.

$$3+3x-2x^{2}-3x=-6$$

Pahekotia te 3x me -3x, ka 0.

$$3-2x^{2}=-6$$

Me tāpiri te 6 ki ngā taha e rua.

$$3-2x^{2}+6=0$$

Tāpirihia te 3 ki te 6, ka 9.

$$9-2x^{2}=0$$

Ko ngā tikanga tātai pūrua pēnei i tēnei nā, me te kīanga tau x^{2} engari kāore he kīanga tau x, ka taea tonu te whakaoti mā te whakamahi i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, ina tuhia ki te tānga ngahuru: ax^{2}+bx+c=0.

$$-2x^{2}+9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -2 mō a, 0 mō b, me 9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-2\right)\times 9}}{2\left(-2\right)}$$

Pūrua 0.

$$x=\frac{0±\sqrt{-4\left(-2\right)\times 9}}{2\left(-2\right)}$$

Whakareatia -4 ki te -2.

$$x=\frac{0±\sqrt{8\times 9}}{2\left(-2\right)}$$

Whakareatia 8 ki te 9.

$$x=\frac{0±\sqrt{72}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te 72.

$$x=\frac{0±6\sqrt{2}}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{0±6\sqrt{2}}{-4}$$

Nā, me whakaoti te whārite x=\frac{0±6\sqrt{2}}{-4} ina he tāpiri te ±.

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

Nā, me whakaoti te whārite x=\frac{0±6\sqrt{2}}{-4} ina he tango te ±.

$$x=\frac{3\sqrt{2}}{2}$$

Kua oti te whārite te whakatau.

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