Whakaoti i te 3x^2-x=?

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

Tauwehe

$$x\left(3x-1\right)$$

Ngā Upane Otinga

Tauwehea te x.

$$x\left(3x-1\right)$$

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.

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

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.

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

Tuhia te pūtakerua o te 1.

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

Ko te tauaro o -1 ko 1.

$$x=\frac{1±1}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{1±1}{6}$$

Nā, me whakaoti te whārite x=\frac{1±1}{6} ina he tāpiri te ±. Tāpiri 1 ki te 1.

$$x=\frac{2}{6}$$

Whakahekea te hautanga \frac{2}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{1}{3}$$

Nā, me whakaoti te whārite x=\frac{1±1}{6} ina he tango te ±. Tango 1 mai i 1.

$$x=\frac{0}{6}$$

Whakawehe 0 ki te 6.

$$x=0$$

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}{3} mō te x_{1} me te 0 mō te x_{2}.

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

Tango \frac{1}{3} 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.

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

Whakakorea atu te tauwehe pūnoa nui rawa 3 i roto i te 3 me te 3.

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

Aromātai

$$x\left(3x-1\right)$$