Whakaoti i te -1+2d^2-d

$$- 1 + 2 d ^ { 2 } - d$$

Tauwehe

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

Ngā Upane Otinga

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$2d^{2}-d-1$$

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2d^{2}+ad+bd-1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

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

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

$$a=-2 b=1$$

Tuhia anō te 2d^{2}-d-1 hei \left(2d^{2}-2d\right)+\left(d-1\right).

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

Whakatauwehea atu 2d i te 2d^{2}-2d.

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

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

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$2d^{2}-d-1=0$$

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.

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

Whakareatia -4 ki te 2.

$$d=\frac{-\left(-1\right)±\sqrt{1-8\left(-1\right)}}{2\times 2}$$

Whakareatia -8 ki te -1.

$$d=\frac{-\left(-1\right)±\sqrt{1+8}}{2\times 2}$$

Tāpiri 1 ki te 8.

$$d=\frac{-\left(-1\right)±\sqrt{9}}{2\times 2}$$

Tuhia te pūtakerua o te 9.

$$d=\frac{-\left(-1\right)±3}{2\times 2}$$

Ko te tauaro o -1 ko 1.

$$d=\frac{1±3}{2\times 2}$$

Whakareatia 2 ki te 2.

$$d=\frac{1±3}{4}$$

Nā, me whakaoti te whārite d=\frac{1±3}{4} ina he tāpiri te ±. Tāpiri 1 ki te 3.

$$d=\frac{4}{4}$$

Whakawehe 4 ki te 4.

$$d=1$$

Nā, me whakaoti te whārite d=\frac{1±3}{4} ina he tango te ±. Tango 3 mai i 1.

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

Whakahekea te hautanga \frac{-2}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$d=-\frac{1}{2}$$

Tauwehea te kīanga taketake mā te whakamahi i te ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Me whakakapi te 1 mō te x_{1} me te -\frac{1}{2} mō te x_{2}.

$$2d^{2}-d-1=2\left(d-1\right)\left(d-\left(-\frac{1}{2}\right)\right)$$

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

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

Tāpiri \frac{1}{2} ki te d mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$2d^{2}-d-1=2\left(d-1\right)\times \frac{2d+1}{2}$$

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 2 me te 2.

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

Aromātai

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