Whakaoti i te 9m^2+m

$$9 m ^ { 2 } + m$$

Tauwehe

$$m\left(9m+1\right)$$

Ngā Upane Otinga

Tauwehea te m.

$$m\left(9m+1\right)$$

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.

$$9m^{2}+m=0$$

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.

$$m=\frac{-1±\sqrt{1^{2}}}{2\times 9}$$

Tuhia te pūtakerua o te 1^{2}.

$$m=\frac{-1±1}{2\times 9}$$

Whakareatia 2 ki te 9.

$$m=\frac{-1±1}{18}$$

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

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

Whakawehe 0 ki te 18.

$$m=0$$

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

$$m=-\frac{2}{18}$$

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

$$m=-\frac{1}{9}$$

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

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

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

$$9m^{2}+m=9m\left(m+\frac{1}{9}\right)$$

Tāpiri \frac{1}{9} ki te m 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.

$$9m^{2}+m=9m\times \frac{9m+1}{9}$$

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

$$9m^{2}+m=m\left(9m+1\right)$$

Aromātai

$$m\left(9m+1\right)$$