Whakaoti i te 3/x=1/x^2*x+4/2x*x

$$\frac { 3 } { x } = \frac { 1 } { x ^ { 2 } } \cdot x + \frac { 4 } { 2 x } \cdot x$$

Whakaoti mō x

$$x=1$$

Ngā Upane Whakamahi Tauwehe

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 2x^{2}, arā, te tauraro pātahi he tino iti rawa te kitea o x,x^{2},2x.

$$2x\times 3=2\times 1x+x\times 4x$$

Whakareatia te x ki te x, ka x^{2}.

$$2x\times 3=2\times 1x+x^{2}\times 4$$

Whakareatia te 2 ki te 3, ka 6.

$$6x=2\times 1x+x^{2}\times 4$$

Whakareatia te 2 ki te 1, ka 2.

$$6x=2x+x^{2}\times 4$$

Tangohia te 2x mai i ngā taha e rua.

$$6x-2x=x^{2}\times 4$$

Pahekotia te 6x me -2x, ka 4x.

$$4x=x^{2}\times 4$$

Tangohia te x^{2}\times 4 mai i ngā taha e rua.

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

Whakareatia te -1 ki te 4, ka -4.

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

Tauwehea te x.

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

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

$$x=0 x=1$$

Tē taea kia ōrite te tāupe x ki 0.

$$x=1$$

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. Me whakarea ngā taha e rua o te whārite ki te 2x^{2}, arā, te tauraro pātahi he tino iti rawa te kitea o x,x^{2},2x.

$$2x\times 3=2\times 1x+x\times 4x$$

Whakareatia te x ki te x, ka x^{2}.

$$2x\times 3=2\times 1x+x^{2}\times 4$$

Whakareatia te 2 ki te 3, ka 6.

$$6x=2\times 1x+x^{2}\times 4$$

Whakareatia te 2 ki te 1, ka 2.

$$6x=2x+x^{2}\times 4$$

Tangohia te 2x mai i ngā taha e rua.

$$6x-2x=x^{2}\times 4$$

Pahekotia te 6x me -2x, ka 4x.

$$4x=x^{2}\times 4$$

Tangohia te x^{2}\times 4 mai i ngā taha e rua.

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

Whakareatia te -1 ki te 4, ka -4.

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

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.

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

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

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

Tuhia te pūtakerua o te 4^{2}.

$$x=\frac{-4±4}{2\left(-4\right)}$$

Whakareatia 2 ki te -4.

$$x=\frac{-4±4}{-8}$$

Nā, me whakaoti te whārite x=\frac{-4±4}{-8} ina he tāpiri te ±. Tāpiri -4 ki te 4.

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

Whakawehe 0 ki te -8.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-4±4}{-8} ina he tango te ±. Tango 4 mai i -4.

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

Whakawehe -8 ki te -8.

$$x=1$$

Kua oti te whārite te whakatau.

$$x=0 x=1$$

Tē taea kia ōrite te tāupe x ki 0.

$$x=1$$

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. Me whakarea ngā taha e rua o te whārite ki te 2x^{2}, arā, te tauraro pātahi he tino iti rawa te kitea o x,x^{2},2x.

$$2x\times 3=2\times 1x+x\times 4x$$

Whakareatia te x ki te x, ka x^{2}.

$$2x\times 3=2\times 1x+x^{2}\times 4$$

Whakareatia te 2 ki te 3, ka 6.

$$6x=2\times 1x+x^{2}\times 4$$

Whakareatia te 2 ki te 1, ka 2.

$$6x=2x+x^{2}\times 4$$

Tangohia te 2x mai i ngā taha e rua.

$$6x-2x=x^{2}\times 4$$

Pahekotia te 6x me -2x, ka 4x.

$$4x=x^{2}\times 4$$

Tangohia te x^{2}\times 4 mai i ngā taha e rua.

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

Whakareatia te -1 ki te 4, ka -4.

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

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.

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

Whakawehea ngā taha e rua ki te -4.

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

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

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

Whakawehe 4 ki te -4.

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

Whakawehe 0 ki te -4.

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

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

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

Pūruatia -\frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tauwehea te x^{2}-x+\frac{1}{4}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{1}{2} x-\frac{1}{2}=-\frac{1}{2}$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=1 x=0$$

Tē taea kia ōrite te tāupe x ki 0.

$$x=1$$