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

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

Whakaoti mō x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x,x+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 2.

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

Pahekotia te 2x me x\times 2, ka 4x.

$$4x+2=3x\left(x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te x+1.

$$4x+2=3x^{2}+3x$$

Tangohia te 3x^{2} mai i ngā taha e rua.

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

Tangohia te 3x mai i ngā taha e rua.

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

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

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

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

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

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

$$a+b=1 ab=-3\times 2=-6$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -6.

$$-1,6 -2,3$$

Tātaihia te tapeke mō ia takirua.

$$-1+6=5 -2+3=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=3 b=-2$$

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

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

Tauwehea te 3x i te tuatahi me te 2 i te rōpū tuarua.

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

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

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

Hei kimi otinga whārite, me whakaoti te -x+1=0 me te 3x+2=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x,x+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 2.

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

Pahekotia te 2x me x\times 2, ka 4x.

$$4x+2=3x\left(x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te x+1.

$$4x+2=3x^{2}+3x$$

Tangohia te 3x^{2} mai i ngā taha e rua.

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

Tangohia te 3x mai i ngā taha e rua.

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

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

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

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.

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

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

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

Pūrua 1.

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

Whakareatia -4 ki te -3.

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

Whakareatia 12 ki te 2.

$$x=\frac{-1±\sqrt{1+24}}{2\left(-3\right)}$$

Tāpiri 1 ki te 24.

$$x=\frac{-1±\sqrt{25}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-1±5}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{-1±5}{-6}$$

Nā, me whakaoti te whārite x=\frac{-1±5}{-6} ina he tāpiri te ±. Tāpiri -1 ki te 5.

$$x=\frac{4}{-6}$$

Whakahekea te hautanga \frac{4}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Nā, me whakaoti te whārite x=\frac{-1±5}{-6} ina he tango te ±. Tango 5 mai i -1.

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

Whakawehe -6 ki te -6.

$$x=1$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x,x+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 2.

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

Pahekotia te 2x me x\times 2, ka 4x.

$$4x+2=3x\left(x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te x+1.

$$4x+2=3x^{2}+3x$$

Tangohia te 3x^{2} mai i ngā taha e rua.

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

Tangohia te 3x mai i ngā taha e rua.

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

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

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

Tangohia te 2 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

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.

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

Whakawehea ngā taha e rua ki te -3.

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

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

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

Whakawehe 1 ki te -3.

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

Whakawehe -2 ki te -3.

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

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.

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

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

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

Tāpiri \frac{2}{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.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{25}{36}$$

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

$$\left(x-\frac{1}{6}\right)^{2}=\frac{25}{36}$$

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

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Whakarūnātia.

$$x-\frac{1}{6}=\frac{5}{6} x-\frac{1}{6}=-\frac{5}{6}$$

Me tāpiri \frac{1}{6} ki ngā taha e rua o te whārite.

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