Whakaoti i te x/x-x-1/3=0.6

$$\frac { x } { x } - \frac { x - 1 } { 3 } = 0.6$$

Whakaoti mō x

$$x=2.2$$

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 3x, arā, te tauraro pātahi he tino iti rawa te kitea o x,3.

$$3x-x\left(x-1\right)=1.8x$$

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

$$3x-\left(x^{2}-x\right)=1.8x$$

Hei kimi i te tauaro o x^{2}-x, kimihia te tauaro o ia taurangi.

$$3x-x^{2}-\left(-x\right)=1.8x$$

Ko te tauaro o -x ko x.

$$3x-x^{2}+x=1.8x$$

Pahekotia te 3x me x, ka 4x.

$$4x-x^{2}=1.8x$$

Tangohia te 1.8x mai i ngā taha e rua.

$$4x-x^{2}-1.8x=0$$

Pahekotia te 4x me -1.8x, ka 2.2x.

$$2.2x-x^{2}=0$$

Tauwehea te x.

$$x\left(2.2-x\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 2.2-x=0.

$$x=0 x=\frac{11}{5}$$

Tē taea kia ōrite te tāupe x ki 0.

$$x=\frac{11}{5}$$

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 3x, arā, te tauraro pātahi he tino iti rawa te kitea o x,3.

$$3x-x\left(x-1\right)=1.8x$$

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

$$3x-\left(x^{2}-x\right)=1.8x$$

Hei kimi i te tauaro o x^{2}-x, kimihia te tauaro o ia taurangi.

$$3x-x^{2}-\left(-x\right)=1.8x$$

Ko te tauaro o -x ko x.

$$3x-x^{2}+x=1.8x$$

Pahekotia te 3x me x, ka 4x.

$$4x-x^{2}=1.8x$$

Tangohia te 1.8x mai i ngā taha e rua.

$$4x-x^{2}-1.8x=0$$

Pahekotia te 4x me -1.8x, ka 2.2x.

$$2.2x-x^{2}=0$$

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.

$$-x^{2}+\frac{11}{5}x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -1 mō a, \frac{11}{5} mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{11}{5}±\sqrt{\left(\frac{11}{5}\right)^{2}}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te \left(\frac{11}{5}\right)^{2}.

$$x=\frac{-\frac{11}{5}±\frac{11}{5}}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-\frac{11}{5}±\frac{11}{5}}{-2}$$

Nā, me whakaoti te whārite x=\frac{-\frac{11}{5}±\frac{11}{5}}{-2} ina he tāpiri te ±. Tāpiri -\frac{11}{5} ki te \frac{11}{5} 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.

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

Whakawehe 0 ki te -2.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-\frac{11}{5}±\frac{11}{5}}{-2} ina he tango te ±. Tango \frac{11}{5} mai i -\frac{11}{5} mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x=-\frac{\frac{22}{5}}{-2}$$

Whakawehe -\frac{22}{5} ki te -2.

$$x=\frac{11}{5}$$

Kua oti te whārite te whakatau.

$$x=0 x=\frac{11}{5}$$

Tē taea kia ōrite te tāupe x ki 0.

$$x=\frac{11}{5}$$

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 3x, arā, te tauraro pātahi he tino iti rawa te kitea o x,3.

$$3x-x\left(x-1\right)=1.8x$$

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

$$3x-\left(x^{2}-x\right)=1.8x$$

Hei kimi i te tauaro o x^{2}-x, kimihia te tauaro o ia taurangi.

$$3x-x^{2}-\left(-x\right)=1.8x$$

Ko te tauaro o -x ko x.

$$3x-x^{2}+x=1.8x$$

Pahekotia te 3x me x, ka 4x.

$$4x-x^{2}=1.8x$$

Tangohia te 1.8x mai i ngā taha e rua.

$$4x-x^{2}-1.8x=0$$

Pahekotia te 4x me -1.8x, ka 2.2x.

$$2.2x-x^{2}=0$$

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.

$$-x^{2}+\frac{11}{5}x=0$$

Whakawehea ngā taha e rua ki te -1.

$$\frac{-x^{2}+\frac{11}{5}x}{-1}=\frac{0}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

$$x^{2}+\frac{\frac{11}{5}}{-1}x=\frac{0}{-1}$$

Whakawehe \frac{11}{5} ki te -1.

$$x^{2}-\frac{11}{5}x=\frac{0}{-1}$$

Whakawehe 0 ki te -1.

$$x^{2}-\frac{11}{5}x=0$$

Whakawehea te -\frac{11}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{11}{10}. Nā, tāpiria te pūrua o te -\frac{11}{10} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{11}{5}x+\left(-\frac{11}{10}\right)^{2}=\left(-\frac{11}{10}\right)^{2}$$

Pūruatia -\frac{11}{10} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=\frac{121}{100}$$

Tauwehea te x^{2}-\frac{11}{5}x+\frac{121}{100}. 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}.

$$\left(x-\frac{11}{10}\right)^{2}=\frac{121}{100}$$

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

$$\sqrt{\left(x-\frac{11}{10}\right)^{2}}=\sqrt{\frac{121}{100}}$$

Whakarūnātia.

$$x-\frac{11}{10}=\frac{11}{10} x-\frac{11}{10}=-\frac{11}{10}$$

Me tāpiri \frac{11}{10} ki ngā taha e rua o te whārite.

$$x=\frac{11}{5} x=0$$

Tē taea kia ōrite te tāupe x ki 0.

$$x=\frac{11}{5}$$