Whakaoti i te 9x^2-x

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

Tauwehe

$$x\left(9x-1\right)$$

Ngā Upane Otinga

Tauwehea te x.

$$x\left(9x-1\right)$$

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.

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

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.

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

Tuhia te pūtakerua o te 1.

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

Ko te tauaro o -1 ko 1.

$$x=\frac{1±1}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{1±1}{18}$$

Nā, me whakaoti te whārite x=\frac{1±1}{18} ina he tāpiri te ±. Tāpiri 1 ki te 1.

$$x=\frac{2}{18}$$

Whakahekea te hautanga \frac{2}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{1}{9}$$

Nā, me whakaoti te whārite x=\frac{1±1}{18} ina he tango te ±. Tango 1 mai i 1.

$$x=\frac{0}{18}$$

Whakawehe 0 ki te 18.

$$x=0$$

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 \frac{1}{9} mō te x_{1} me te 0 mō te x_{2}.

$$9x^{2}-x=9\left(x-\frac{1}{9}\right)x$$

Tango \frac{1}{9} mai i x mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$9x^{2}-x=9\times \frac{9x-1}{9}x$$

Whakakorea atu te tauwehe pūnoa nui rawa 9 i roto i te 9 me te 9.

$$9x^{2}-x=\left(9x-1\right)x$$

Aromātai

$$x\left(9x-1\right)$$