Whakaoti i te 1-2x/x+7=x/x-1

$$\frac { 1 - 2 x } { x + 7 } = \frac { x } { x - 1 }$$

Whakaoti mō x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -7,1 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(x+7\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+7,x-1.

$$\left(x-1\right)\left(1-2x\right)=\left(x+7\right)x$$

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

$$3x-2x^{2}-1=\left(x+7\right)x$$

Whakamahia te āhuatanga tohatoha hei whakarea te x+7 ki te x.

$$3x-2x^{2}-1=x^{2}+7x$$

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

$$3x-2x^{2}-1-x^{2}=7x$$

Pahekotia te -2x^{2} me -x^{2}, ka -3x^{2}.

$$3x-3x^{2}-1=7x$$

Tangohia te 7x mai i ngā taha e rua.

$$3x-3x^{2}-1-7x=0$$

Pahekotia te 3x me -7x, ka -4x.

$$-4x-3x^{2}-1=0$$

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$-3x^{2}-4x-1=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -3x^{2}+ax+bx-1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=-3\left(-1\right)=3$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-1 b=-3$$

Tuhia anō te -3x^{2}-4x-1 hei \left(-3x^{2}-x\right)+\left(-3x-1\right).

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

Tauwehea te -x i te tuatahi me te -1 i te rōpū tuarua.

$$-x\left(3x+1\right)-\left(3x+1\right)$$

Whakatauwehea atu te kīanga pātahi 3x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x+1\right)\left(-x-1\right)$$

Hei kimi otinga whārite, me whakaoti te 3x+1=0 me te -x-1=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -7,1 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(x+7\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+7,x-1.

$$\left(x-1\right)\left(1-2x\right)=\left(x+7\right)x$$

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

$$3x-2x^{2}-1=\left(x+7\right)x$$

Whakamahia te āhuatanga tohatoha hei whakarea te x+7 ki te x.

$$3x-2x^{2}-1=x^{2}+7x$$

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

$$3x-2x^{2}-1-x^{2}=7x$$

Pahekotia te -2x^{2} me -x^{2}, ka -3x^{2}.

$$3x-3x^{2}-1=7x$$

Tangohia te 7x mai i ngā taha e rua.

$$3x-3x^{2}-1-7x=0$$

Pahekotia te 3x me -7x, ka -4x.

$$-4x-3x^{2}-1=0$$

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.

$$-3x^{2}-4x-1=0$$

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

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

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-3\right)\left(-1\right)}}{2\left(-3\right)}$$

Whakareatia -4 ki te -3.

$$x=\frac{-\left(-4\right)±\sqrt{16+12\left(-1\right)}}{2\left(-3\right)}$$

Whakareatia 12 ki te -1.

$$x=\frac{-\left(-4\right)±\sqrt{16-12}}{2\left(-3\right)}$$

Tāpiri 16 ki te -12.

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

Tuhia te pūtakerua o te 4.

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

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{4±2}{-6}$$

Nā, me whakaoti te whārite x=\frac{4±2}{-6} ina he tāpiri te ±. Tāpiri 4 ki te 2.

$$x=\frac{6}{-6}$$

Whakawehe 6 ki te -6.

$$x=-1$$

Nā, me whakaoti te whārite x=\frac{4±2}{-6} ina he tango te ±. Tango 2 mai i 4.

$$x=\frac{2}{-6}$$

Whakahekea te hautanga \frac{2}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -7,1 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(x+7\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+7,x-1.

$$\left(x-1\right)\left(1-2x\right)=\left(x+7\right)x$$

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

$$3x-2x^{2}-1=\left(x+7\right)x$$

Whakamahia te āhuatanga tohatoha hei whakarea te x+7 ki te x.

$$3x-2x^{2}-1=x^{2}+7x$$

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

$$3x-2x^{2}-1-x^{2}=7x$$

Pahekotia te -2x^{2} me -x^{2}, ka -3x^{2}.

$$3x-3x^{2}-1=7x$$

Tangohia te 7x mai i ngā taha e rua.

$$3x-3x^{2}-1-7x=0$$

Pahekotia te 3x me -7x, ka -4x.

$$-4x-3x^{2}-1=0$$

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

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

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.

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

Whakawehea ngā taha e rua ki te -3.

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

Mā te whakawehe ki te -3 ka wetekia te whakareanga ki te -3.

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

Whakawehe -4 ki te -3.

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

Whakawehe 1 ki te -3.

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

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

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

Pūruatia \frac{2}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=-\frac{1}{3}+\frac{4}{9}$$

Tāpiri -\frac{1}{3} ki te \frac{4}{9} 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.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{1}{9}$$

Tauwehea te x^{2}+\frac{4}{3}x+\frac{4}{9}. 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}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{1}{9}$$

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

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Whakarūnātia.

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

Me tango \frac{2}{3} mai i ngā taha e rua o te whārite.

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