Whakaoti i te 3(1-x)+4(1+2x)(1-x)=7

$$3(1-x)+4(1+2x)(1-x)=7$$

Whakaoti mō x

$$x=\frac{1}{8}=0.125$$

$$x=0$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te 1-x.

$$3-3x+4\left(1+2x\right)\left(1-x\right)=7$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te 1+2x.

$$3-3x+\left(4+8x\right)\left(1-x\right)=7$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 4+8x ki te 1-x ka whakakotahi i ngā kupu rite.

$$3-3x+4+4x-8x^{2}=7$$

Tāpirihia te 3 ki te 4, ka 7.

$$7-3x+4x-8x^{2}=7$$

Pahekotia te -3x me 4x, ka x.

$$7+x-8x^{2}=7$$

Tangohia te 7 mai i ngā taha e rua.

$$7+x-8x^{2}-7=0$$

Tangohia te 7 i te 7, ka 0.

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

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -8 mō a, 1 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}}}{2\left(-8\right)}$$

Tuhia te pūtakerua o te 1^{2}.

$$x=\frac{-1±1}{2\left(-8\right)}$$

Whakareatia 2 ki te -8.

$$x=\frac{-1±1}{-16}$$

Nā, me whakaoti te whārite x=\frac{-1±1}{-16} ina he tāpiri te ±. Tāpiri -1 ki te 1.

$$x=\frac{0}{-16}$$

Whakawehe 0 ki te -16.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-1±1}{-16} ina he tango te ±. Tango 1 mai i -1.

$$x=-\frac{2}{-16}$$

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

$$x=\frac{1}{8}$$

Kua oti te whārite te whakatau.

$$x=0 x=\frac{1}{8}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te 1-x.

$$3-3x+4\left(1+2x\right)\left(1-x\right)=7$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te 1+2x.

$$3-3x+\left(4+8x\right)\left(1-x\right)=7$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 4+8x ki te 1-x ka whakakotahi i ngā kupu rite.

$$3-3x+4+4x-8x^{2}=7$$

Tāpirihia te 3 ki te 4, ka 7.

$$7-3x+4x-8x^{2}=7$$

Pahekotia te -3x me 4x, ka x.

$$7+x-8x^{2}=7$$

Tangohia te 7 mai i ngā taha e rua.

$$x-8x^{2}=7-7$$

Tangohia te 7 i te 7, ka 0.

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

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.

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

Whakawehea ngā taha e rua ki te -8.

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

Mā te whakawehe ki te -8 ka wetekia te whakareanga ki te -8.

$$x^{2}+\frac{1}{-8}x=\frac{0}{-8}$$

Whakawehe 1 ki te -8.

$$x^{2}-\frac{1}{8}x=\frac{0}{-8}$$

Whakawehe 0 ki te -8.

$$x^{2}-\frac{1}{8}x=0$$

Whakawehea te -\frac{1}{8}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{16}. Nā, tāpiria te pūrua o te -\frac{1}{16} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{1}{8}x+\left(-\frac{1}{16}\right)^{2}=\left(-\frac{1}{16}\right)^{2}$$

Pūruatia -\frac{1}{16} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{8}x+\frac{1}{256}=\frac{1}{256}$$

Tauwehea x^{2}-\frac{1}{8}x+\frac{1}{256}. 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}.

$$\left(x-\frac{1}{16}\right)^{2}=\frac{1}{256}$$

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

$$\sqrt{\left(x-\frac{1}{16}\right)^{2}}=\sqrt{\frac{1}{256}}$$

Whakarūnātia.

$$x-\frac{1}{16}=\frac{1}{16} x-\frac{1}{16}=-\frac{1}{16}$$

Me tāpiri \frac{1}{16} ki ngā taha e rua o te whārite.

$$x=\frac{1}{8} x=0$$