Whakaoti i te x-1/2x+1=2x+1/x-1+3

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

Whakaoti mō x

$$x=\frac{\sqrt{13}-1}{6}\approx 0.434258546$$

$$x=\frac{-\sqrt{13}-1}{6}\approx -0.767591879$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -\frac{1}{2},1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-1\right)\left(2x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2x+1,x-1.

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

Whakareatia te x-1 ki te x-1, ka \left(x-1\right)^{2}.

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

Whakareatia te 2x+1 ki te 2x+1, ka \left(2x+1\right)^{2}.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-1\right)^{2}.

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x-1 ki te 2x+1 ka whakakotahi i ngā kupu rite.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2x^{2}-x-1 ki te 3.

$$x^{2}-2x+1=4x^{2}+4x+1+6x^{2}-3x-3$$

Pahekotia te 4x^{2} me 6x^{2}, ka 10x^{2}.

$$x^{2}-2x+1=10x^{2}+4x+1-3x-3$$

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

$$x^{2}-2x+1=10x^{2}+x+1-3$$

Tangohia te 3 i te 1, ka -2.

$$x^{2}-2x+1=10x^{2}+x-2$$

Tangohia te 10x^{2} mai i ngā taha e rua.

$$x^{2}-2x+1-10x^{2}=x-2$$

Pahekotia te x^{2} me -10x^{2}, ka -9x^{2}.

$$-9x^{2}-2x+1=x-2$$

Tangohia te x mai i ngā taha e rua.

$$-9x^{2}-2x+1-x=-2$$

Pahekotia te -2x me -x, ka -3x.

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

Me tāpiri te 2 ki ngā taha e rua.

$$-9x^{2}-3x+1+2=0$$

Tāpirihia te 1 ki te 2, ka 3.

$$-9x^{2}-3x+3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -9 mō a, -3 mō b, me 3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -3.

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

Whakareatia -4 ki te -9.

$$x=\frac{-\left(-3\right)±\sqrt{9+36\times 3}}{2\left(-9\right)}$$

Whakareatia 36 ki te 3.

$$x=\frac{-\left(-3\right)±\sqrt{9+108}}{2\left(-9\right)}$$

Tāpiri 9 ki te 108.

$$x=\frac{-\left(-3\right)±\sqrt{117}}{2\left(-9\right)}$$

Tuhia te pūtakerua o te 117.

$$x=\frac{-\left(-3\right)±3\sqrt{13}}{2\left(-9\right)}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±3\sqrt{13}}{2\left(-9\right)}$$

Whakareatia 2 ki te -9.

$$x=\frac{3±3\sqrt{13}}{-18}$$

Nā, me whakaoti te whārite x=\frac{3±3\sqrt{13}}{-18} ina he tāpiri te ±. Tāpiri 3 ki te 3\sqrt{13}.

$$x=\frac{3\sqrt{13}+3}{-18}$$

Whakawehe 3+3\sqrt{13} ki te -18.

$$x=\frac{-\sqrt{13}-1}{6}$$

Nā, me whakaoti te whārite x=\frac{3±3\sqrt{13}}{-18} ina he tango te ±. Tango 3\sqrt{13} mai i 3.

$$x=\frac{3-3\sqrt{13}}{-18}$$

Whakawehe 3-3\sqrt{13} ki te -18.

$$x=\frac{\sqrt{13}-1}{6}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{13}-1}{6} x=\frac{\sqrt{13}-1}{6}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -\frac{1}{2},1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-1\right)\left(2x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2x+1,x-1.

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

Whakareatia te x-1 ki te x-1, ka \left(x-1\right)^{2}.

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

Whakareatia te 2x+1 ki te 2x+1, ka \left(2x+1\right)^{2}.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-1\right)^{2}.

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x-1 ki te 2x+1 ka whakakotahi i ngā kupu rite.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2x^{2}-x-1 ki te 3.

$$x^{2}-2x+1=4x^{2}+4x+1+6x^{2}-3x-3$$

Pahekotia te 4x^{2} me 6x^{2}, ka 10x^{2}.

$$x^{2}-2x+1=10x^{2}+4x+1-3x-3$$

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

$$x^{2}-2x+1=10x^{2}+x+1-3$$

Tangohia te 3 i te 1, ka -2.

$$x^{2}-2x+1=10x^{2}+x-2$$

Tangohia te 10x^{2} mai i ngā taha e rua.

$$x^{2}-2x+1-10x^{2}=x-2$$

Pahekotia te x^{2} me -10x^{2}, ka -9x^{2}.

$$-9x^{2}-2x+1=x-2$$

Tangohia te x mai i ngā taha e rua.

$$-9x^{2}-2x+1-x=-2$$

Pahekotia te -2x me -x, ka -3x.

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

Tangohia te 1 mai i ngā taha e rua.

$$-9x^{2}-3x=-2-1$$

Tangohia te 1 i te -2, ka -3.

$$-9x^{2}-3x=-3$$

Whakawehea ngā taha e rua ki te -9.

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

Mā te whakawehe ki te -9 ka wetekia te whakareanga ki te -9.

$$x^{2}+\left(-\frac{3}{-9}\right)x=-\frac{3}{-9}$$

Whakahekea te hautanga \frac{-3}{-9} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

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

Whakahekea te hautanga \frac{-3}{-9} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

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

Whakawehea te \frac{1}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{6}. Nā, tāpiria te pūrua o te \frac{1}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=\frac{1}{3}+\left(\frac{1}{6}\right)^{2}$$

Pūruatia \frac{1}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{1}{3}+\frac{1}{36}$$

Tāpiri \frac{1}{3} ki te \frac{1}{36} 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.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{13}{36}$$

Tauwehea te x^{2}+\frac{1}{3}x+\frac{1}{36}. 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}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{13}{36}$$

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

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{13}{36}}$$

Whakarūnātia.

$$x+\frac{1}{6}=\frac{\sqrt{13}}{6} x+\frac{1}{6}=-\frac{\sqrt{13}}{6}$$

Me tango \frac{1}{6} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{13}-1}{6} x=\frac{-\sqrt{13}-1}{6}$$