Whakaoti i te x^3-1/(x-1)^2-x^3+1/(x+1)^2=6

$$\frac { x ^ { 3 } - 1 } { ( x - 1 ) ^ { 2 } } - \frac { x ^ { 3 } + 1 } { ( x + 1 ) ^ { 2 } } = 6$$

Whakaoti mō x

$$x=2$$

$$x=-2$$

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

$$\left(x+1\right)^{2}\left(x^{3}-1\right)-\left(x-1\right)^{2}\left(x^{3}+1\right)=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

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

$$\left(x^{2}+2x+1\right)\left(x^{3}-1\right)-\left(x-1\right)^{2}\left(x^{3}+1\right)=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}+2x+1 ki te x^{3}-1.

$$x^{5}-x^{2}+2x^{4}-2x+x^{3}-1-\left(x-1\right)^{2}\left(x^{3}+1\right)=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

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

$$x^{5}-x^{2}+2x^{4}-2x+x^{3}-1-\left(x^{2}-2x+1\right)\left(x^{3}+1\right)=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2}-2x+1 ki te x^{3}+1.

$$x^{5}-x^{2}+2x^{4}-2x+x^{3}-1-\left(x^{5}+x^{2}-2x^{4}-2x+x^{3}+1\right)=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

Hei kimi i te tauaro o x^{5}+x^{2}-2x^{4}-2x+x^{3}+1, kimihia te tauaro o ia taurangi.

$$x^{5}-x^{2}+2x^{4}-2x+x^{3}-1-x^{5}-x^{2}+2x^{4}+2x-x^{3}-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

Pahekotia te x^{5} me -x^{5}, ka 0.

$$-x^{2}+2x^{4}-2x+x^{3}-1-x^{2}+2x^{4}+2x-x^{3}-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

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

$$-2x^{2}+2x^{4}-2x+x^{3}-1+2x^{4}+2x-x^{3}-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

Pahekotia te 2x^{4} me 2x^{4}, ka 4x^{4}.

$$-2x^{2}+4x^{4}-2x+x^{3}-1+2x-x^{3}-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

Pahekotia te -2x me 2x, ka 0.

$$-2x^{2}+4x^{4}+x^{3}-1-x^{3}-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

Pahekotia te x^{3} me -x^{3}, ka 0.

$$-2x^{2}+4x^{4}-1-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

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

$$-2x^{2}+4x^{4}-2=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

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

$$-2x^{2}+4x^{4}-2=6\left(x^{2}-2x+1\right)\left(x+1\right)^{2}$$

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

$$-2x^{2}+4x^{4}-2=6\left(x^{2}-2x+1\right)\left(x^{2}+2x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 6 ki te x^{2}-2x+1.

$$-2x^{2}+4x^{4}-2=\left(6x^{2}-12x+6\right)\left(x^{2}+2x+1\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 6x^{2}-12x+6 ki te x^{2}+2x+1 ka whakakotahi i ngā kupu rite.

$$-2x^{2}+4x^{4}-2=6x^{4}-12x^{2}+6$$

Tangohia te 6x^{4} mai i ngā taha e rua.

$$-2x^{2}+4x^{4}-2-6x^{4}=-12x^{2}+6$$

Pahekotia te 4x^{4} me -6x^{4}, ka -2x^{4}.

$$-2x^{2}-2x^{4}-2=-12x^{2}+6$$

Me tāpiri te 12x^{2} ki ngā taha e rua.

$$-2x^{2}-2x^{4}-2+12x^{2}=6$$

Pahekotia te -2x^{2} me 12x^{2}, ka 10x^{2}.

$$10x^{2}-2x^{4}-2=6$$

Tangohia te 6 mai i ngā taha e rua.

$$10x^{2}-2x^{4}-2-6=0$$

Tangohia te 6 i te -2, ka -8.

$$10x^{2}-2x^{4}-8=0$$

Whakakapia te t mō te x^{2}.

$$-2t^{2}+10t-8=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te -2 mō te a, te 10 mō te b, me te -8 mō te c i te ture pūrua.

$$t=\frac{-10±\sqrt{10^{2}-4\left(-2\right)\left(-8\right)}}{-2\times 2}$$

Mahia ngā tātaitai.

$$t=\frac{-10±6}{-4}$$

Whakaotia te whārite t=\frac{-10±6}{-4} ina he tōrunga te ±, ina he tōraro te ±.

$$t=1 t=4$$

I te mea ko x=t^{2}, ka riro ngā otinga mā te arotake i te x=±\sqrt{t} mō ia t.

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

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara 1,-1.

$$x=-2 x=2$$