Whakaoti i te 9(x^2)=3x

$$9 ( x ^ { 2 } ) = 3 x$$

Whakaoti mō x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tangohia te 3x mai i ngā taha e rua.

$$9x^{2}-3x=0$$

Tauwehea te x.

$$x\left(9x-3\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 9x-3=0.

$$x=0 x=\frac{1}{3}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 3x mai i ngā taha e rua.

$$9x^{2}-3x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 9 mō a, -3 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\times 9}$$

Tuhia te pūtakerua o te \left(-3\right)^{2}.

$$x=\frac{-\left(-3\right)±3}{2\times 9}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±3}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{3±3}{18}$$

Nā, me whakaoti te whārite x=\frac{3±3}{18} ina he tāpiri te ±. Tāpiri 3 ki te 3.

$$x=\frac{6}{18}$$

Whakahekea te hautanga \frac{6}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

Nā, me whakaoti te whārite x=\frac{3±3}{18} ina he tango te ±. Tango 3 mai i 3.

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

Whakawehe 0 ki te 18.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=\frac{1}{3} x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 3x mai i ngā taha e rua.

$$9x^{2}-3x=0$$

Whakawehea ngā taha e rua ki te 9.

$$\frac{9x^{2}-3x}{9}=\frac{0}{9}$$

Mā te whakawehe ki te 9 ka wetekia te whakareanga ki te 9.

$$x^{2}+\left(-\frac{3}{9}\right)x=\frac{0}{9}$$

Whakahekea te hautanga \frac{-3}{9} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$x^{2}-\frac{1}{3}x=\frac{0}{9}$$

Whakawehe 0 ki te 9.

$$x^{2}-\frac{1}{3}x=0$$

Whakawehea te -\frac{1}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{6}. Nā, tāpiria te pūrua o te -\frac{1}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{1}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{1}{36}$$

Tauwehea x^{2}-\frac{1}{3}x+\frac{1}{36}. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{1}{36}$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Whakarūnātia.

$$x-\frac{1}{6}=\frac{1}{6} x-\frac{1}{6}=-\frac{1}{6}$$

Me tāpiri \frac{1}{6} ki ngā taha e rua o te whārite.

$$x=\frac{1}{3} x=0$$