Whakaoti i te x+1=3x^2+1+0

$$x + 1 = 3 x ^ { 2 } + 1 + 0$$

Whakaoti mō x

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

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tāpirihia te 1 ki te 0, ka 1.

$$x+1=3x^{2}+1$$

Tangohia te 3x^{2} mai i ngā taha e rua.

$$x+1-3x^{2}=1$$

Tangohia te 1 mai i ngā taha e rua.

$$x+1-3x^{2}-1=0$$

Tangohia te 1 i te 1, ka 0.

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

Tauwehea te x.

$$x\left(1-3x\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 1-3x=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tāpirihia te 1 ki te 0, ka 1.

$$x+1=3x^{2}+1$$

Tangohia te 3x^{2} mai i ngā taha e rua.

$$x+1-3x^{2}=1$$

Tangohia te 1 mai i ngā taha e rua.

$$x+1-3x^{2}-1=0$$

Tangohia te 1 i te 1, ka 0.

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

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.

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

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

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

Tuhia te pūtakerua o te 1^{2}.

$$x=\frac{-1±1}{2\left(-3\right)}$$

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.

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

Whakawehe 0 ki te -6.

$$x=0$$

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

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tāpirihia te 1 ki te 0, ka 1.

$$x+1=3x^{2}+1$$

Tangohia te 3x^{2} mai i ngā taha e rua.

$$x+1-3x^{2}=1$$

Tangohia te 1 mai i ngā taha e rua.

$$x-3x^{2}=1-1$$

Tangohia te 1 i te 1, ka 0.

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

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

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

Whakawehea ngā taha e rua ki te -3.

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

Mā te whakawehe ki te -3 ka wetekia te whakareanga ki te -3.

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

Whakawehe 1 ki te -3.

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

Whakawehe 0 ki te -3.

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

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.

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