Whakaoti i te x^2-3x/3x^2+16x-2=0

$$\frac{ { x }^{ 2 } -3x }{ 3 { x }^{ 2 } +16x-2 } =0$$

Whakaoti mō x

$$x=3$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Whakareatia ngā taha e rua o te whārite ki te 3\left(x-\left(-\frac{1}{3}\sqrt{70}-\frac{8}{3}\right)\right)\left(x-\left(\frac{1}{3}\sqrt{70}-\frac{8}{3}\right)\right).

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

Tauwehea te x.

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

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

$$x=0 x=3$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia ngā taha e rua o te whārite ki te 3\left(x-\left(-\frac{1}{3}\sqrt{70}-\frac{8}{3}\right)\right)\left(x-\left(\frac{1}{3}\sqrt{70}-\frac{8}{3}\right)\right).

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -3 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

$$x=\frac{-\left(-3\right)±3}{2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±3}{2}$$

Nā, me whakaoti te whārite x=\frac{3±3}{2} ina he tāpiri te ±. Tāpiri 3 ki te 3.

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

Whakawehe 6 ki te 2.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{3±3}{2} ina he tango te ±. Tango 3 mai i 3.

$$x=\frac{0}{2}$$

Whakawehe 0 ki te 2.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=3 x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia ngā taha e rua o te whārite ki te 3\left(x-\left(-\frac{1}{3}\sqrt{70}-\frac{8}{3}\right)\right)\left(x-\left(\frac{1}{3}\sqrt{70}-\frac{8}{3}\right)\right).

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

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

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

Pūruatia -\frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-3x+\frac{9}{4}=\frac{9}{4}$$

Tauwehea x^{2}-3x+\frac{9}{4}. 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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{9}{4}$$

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

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Whakarūnātia.

$$x-\frac{3}{2}=\frac{3}{2} x-\frac{3}{2}=-\frac{3}{2}$$

Me tāpiri \frac{3}{2} ki ngā taha e rua o te whārite.

$$x=3 x=0$$