Whakaoti i te 4x^2+24x/32x=1

$$\frac { 4 x ^ { 2 } + 24 x } { 32 x } = 1$$

Whakaoti mō x

$$x=2$$

Ngā Upane Whakamahi Tauwehe

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te 32x.

$$4x^{2}+24x=32x$$

Tangohia te 32x mai i ngā taha e rua.

$$4x^{2}+24x-32x=0$$

Pahekotia te 24x me -32x, ka -8x.

$$4x^{2}-8x=0$$

Tauwehea te x.

$$x\left(4x-8\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 4x-8=0.

$$x=0 x=2$$

Tē taea kia ōrite te tāupe x ki 0.

$$x=2$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te 32x.

$$4x^{2}+24x=32x$$

Tangohia te 32x mai i ngā taha e rua.

$$4x^{2}+24x-32x=0$$

Pahekotia te 24x me -32x, ka -8x.

$$4x^{2}-8x=0$$

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

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 4}$$

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

$$x=\frac{-\left(-8\right)±8}{2\times 4}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8±8}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{8±8}{8}$$

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

$$x=\frac{16}{8}$$

Whakawehe 16 ki te 8.

$$x=2$$

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

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

Whakawehe 0 ki te 8.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=2 x=0$$

Tē taea kia ōrite te tāupe x ki 0.

$$x=2$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te 32x.

$$4x^{2}+24x=32x$$

Tangohia te 32x mai i ngā taha e rua.

$$4x^{2}+24x-32x=0$$

Pahekotia te 24x me -32x, ka -8x.

$$4x^{2}-8x=0$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}-8x}{4}=\frac{0}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

$$x^{2}+\left(-\frac{8}{4}\right)x=\frac{0}{4}$$

Whakawehe -8 ki te 4.

$$x^{2}-2x=\frac{0}{4}$$

Whakawehe 0 ki te 4.

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

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

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

Tauwehea x^{2}-2x+1. 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}.

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

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

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

Whakarūnātia.

$$x-1=1 x-1=-1$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$x=2 x=0$$

Tē taea kia ōrite te tāupe x ki 0.

$$x=2$$