Whakaoti i te 3(3x(1/3+x))=9x-1

$$3(3x( \frac{ 1 }{ 3 } +x))=9x-1$$

Whakaoti mō x

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia te 3 ki te 3, ka 9.

$$9x\left(\frac{1}{3}+x\right)=9x-1$$

Whakamahia te āhuatanga tohatoha hei whakarea te 9x ki te \frac{1}{3}+x.

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

Whakareatia te 9 ki te \frac{1}{3}, ka \frac{9}{3}.

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

Whakawehea te 9 ki te 3, kia riro ko 3.

$$3x+9x^{2}=9x-1$$

Tangohia te 9x mai i ngā taha e rua.

$$3x+9x^{2}-9x=-1$$

Pahekotia te 3x me -9x, ka -6x.

$$-6x+9x^{2}=-1$$

Me tāpiri te 1 ki ngā taha e rua.

$$-6x+9x^{2}+1=0$$

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.

$$9x^{2}-6x+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 9 mō a, -6 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9}}{2\times 9}$$

Pūrua -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 9}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$x=\frac{-\left(-6\right)±\sqrt{36-36}}{2\times 9}$$

Tāpiri 36 ki te -36.

$$x=\frac{-\left(-6\right)±\sqrt{0}}{2\times 9}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{-6}{2\times 9}$$

Ko te tauaro o -6 ko 6.

$$x=\frac{6}{2\times 9}$$

Whakareatia 2 ki te 9.

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

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

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia te 3 ki te 3, ka 9.

$$9x\left(\frac{1}{3}+x\right)=9x-1$$

Whakamahia te āhuatanga tohatoha hei whakarea te 9x ki te \frac{1}{3}+x.

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

Whakareatia te 9 ki te \frac{1}{3}, ka \frac{9}{3}.

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

Whakawehea te 9 ki te 3, kia riro ko 3.

$$3x+9x^{2}=9x-1$$

Tangohia te 9x mai i ngā taha e rua.

$$3x+9x^{2}-9x=-1$$

Pahekotia te 3x me -9x, ka -6x.

$$-6x+9x^{2}=-1$$

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.

$$9x^{2}-6x=-1$$

Whakawehea ngā taha e rua ki te 9.

$$\frac{9x^{2}-6x}{9}=-\frac{1}{9}$$

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

$$x^{2}+\left(-\frac{6}{9}\right)x=-\frac{1}{9}$$

Whakahekea te hautanga \frac{-6}{9} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

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

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

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=-\frac{1}{9}+\left(-\frac{1}{3}\right)^{2}$$

Pūruatia -\frac{1}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri -\frac{1}{9} ki te \frac{1}{9} 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.

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

Tauwehea x^{2}-\frac{2}{3}x+\frac{1}{9}. 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}.

$$\left(x-\frac{1}{3}\right)^{2}=0$$

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

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

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

Me tāpiri \frac{1}{3} ki ngā taha e rua o te whārite.

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

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

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