Whakaoti i te 2x-1/x=3

$$2x- \frac{ 1 }{ x } = 3$$

Whakaoti mō x

$$x = \frac{\sqrt{17} + 3}{4} \approx 1.780776406$$

$$x=\frac{3-\sqrt{17}}{4}\approx -0.280776406$$

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 x.

$$2xx-1=3x$$

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

$$2x^{2}-1=3x$$

Tangohia te 3x mai i ngā taha e rua.

$$2x^{2}-1-3x=0$$

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.

$$2x^{2}-3x-1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -3 mō b, me -1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-1\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-1\right)}}{2\times 2}$$

Whakareatia -8 ki te -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+8}}{2\times 2}$$

Tāpiri 9 ki te 8.

$$x=\frac{-\left(-3\right)±\sqrt{17}}{2\times 2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±\sqrt{17}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{3±\sqrt{17}}{4}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{17}}{4} ina he tāpiri te ±. Tāpiri 3 ki te \sqrt{17}.

$$x=\frac{\sqrt{17}+3}{4}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{17}}{4} ina he tango te ±. Tango \sqrt{17} mai i 3.

$$x=\frac{3-\sqrt{17}}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{17}+3}{4} x=\frac{3-\sqrt{17}}{4}$$

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 x.

$$2xx-1=3x$$

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

$$2x^{2}-1=3x$$

Tangohia te 3x mai i ngā taha e rua.

$$2x^{2}-1-3x=0$$

Me tāpiri te 1 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$2x^{2}-3x=1$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}-3x}{2}=\frac{1}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$x^{2}-\frac{3}{2}x=\frac{1}{2}$$

Whakawehea te -\frac{3}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{3}{4}. Nā, tāpiria te pūrua o te -\frac{3}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{1}{2}+\left(-\frac{3}{4}\right)^{2}$$

Pūruatia -\frac{3}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{2}+\frac{9}{16}$$

Tāpiri \frac{1}{2} ki te \frac{9}{16} 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.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{17}{16}$$

Tauwehea x^{2}-\frac{3}{2}x+\frac{9}{16}. 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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{17}{16}$$

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

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Whakarūnātia.

$$x-\frac{3}{4}=\frac{\sqrt{17}}{4} x-\frac{3}{4}=-\frac{\sqrt{17}}{4}$$

Me tāpiri \frac{3}{4} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{17}+3}{4} x=\frac{3-\sqrt{17}}{4}$$