Whakaoti i te {l}{x=3(3x(1-x))(1-3x(1-x)}{=9x(1-x)(1-(3x-3x^2)))}

$$\left. \begin{array} { l } { x = 3 ( 3 x ( 1 - x ) ) ( 1 - 3 x ( 1 - x ) } \\ { = 9 x ( 1 - x ) ( 1 - ( 3 x - 3 x ^ { 2 } ) ) ) } \end{array} \right.$$

Whakaoti mō x

$$x=\frac{8}{9}\approx 0.888888889$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Whakareatia te 3 ki te 3, ka 9.

$$x=9x\left(1-x\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 9x ki te 1-x.

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

Tangohia te 9x mai i ngā taha e rua.

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

Pahekotia te x me -9x, ka -8x.

$$-8x=-9x^{2}$$

Me tāpiri te 9x^{2} ki ngā taha e rua.

$$-8x+9x^{2}=0$$

Tauwehea te x.

$$x\left(-8+9x\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te -8+9x=0.

$$x=0 x=\frac{8}{9}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia te 3 ki te 3, ka 9.

$$x=9x\left(1-x\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 9x ki te 1-x.

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

Tangohia te 9x mai i ngā taha e rua.

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

Pahekotia te x me -9x, ka -8x.

$$-8x=-9x^{2}$$

Me tāpiri te 9x^{2} ki ngā taha e rua.

$$-8x+9x^{2}=0$$

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.

$$9x^{2}-8x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 9 mō a, -8 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 9}$$

Tuhia te pūtakerua o te \left(-8\right)^{2}.

$$x=\frac{-\left(-8\right)±8}{2\times 9}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8±8}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{8±8}{18}$$

Nā, me whakaoti te whārite x=\frac{8±8}{18} ina he tāpiri te ±. Tāpiri 8 ki te 8.

$$x=\frac{16}{18}$$

Whakahekea te hautanga \frac{16}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{8}{9}$$

Nā, me whakaoti te whārite x=\frac{8±8}{18} ina he tango te ±. Tango 8 mai i 8.

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

Whakawehe 0 ki te 18.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=\frac{8}{9} x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia te 3 ki te 3, ka 9.

$$x=9x\left(1-x\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 9x ki te 1-x.

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

Tangohia te 9x mai i ngā taha e rua.

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

Pahekotia te x me -9x, ka -8x.

$$-8x=-9x^{2}$$

Me tāpiri te 9x^{2} ki ngā taha e rua.

$$-8x+9x^{2}=0$$

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.

$$9x^{2}-8x=0$$

Whakawehea ngā taha e rua ki te 9.

$$\frac{9x^{2}-8x}{9}=\frac{0}{9}$$

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

$$x^{2}-\frac{8}{9}x=\frac{0}{9}$$

Whakawehe 0 ki te 9.

$$x^{2}-\frac{8}{9}x=0$$

Whakawehea te -\frac{8}{9}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{4}{9}. Nā, tāpiria te pūrua o te -\frac{4}{9} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{8}{9}x+\left(-\frac{4}{9}\right)^{2}=\left(-\frac{4}{9}\right)^{2}$$

Pūruatia -\frac{4}{9} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{8}{9}x+\frac{16}{81}=\frac{16}{81}$$

Tauwehea x^{2}-\frac{8}{9}x+\frac{16}{81}. 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}.

$$\left(x-\frac{4}{9}\right)^{2}=\frac{16}{81}$$

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

$$\sqrt{\left(x-\frac{4}{9}\right)^{2}}=\sqrt{\frac{16}{81}}$$

Whakarūnātia.

$$x-\frac{4}{9}=\frac{4}{9} x-\frac{4}{9}=-\frac{4}{9}$$

Me tāpiri \frac{4}{9} ki ngā taha e rua o te whārite.

$$x=\frac{8}{9} x=0$$