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

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

Whakaoti mō x

$$x=-2$$

Ngā Upane Whakamahi Tauwehe

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

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

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

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

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

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

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

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

Pahekotia te 3x me x, ka 4x.

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

Tangohia te x mai i ngā taha e rua.

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

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

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

Me tāpiri te 2 ki ngā taha e rua.

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

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

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

Hei whakaoti i te whārite, whakatauwehea te x^{2}+3x+2 mā te whakamahi i te tātai x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=3 ab=2$$

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

$$a=1 b=2$$

Me tuhi anō te kīanga whakatauwehe \left(x+a\right)\left(x+b\right) mā ngā uara i tātaihia.

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

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

$$x=-1 x=-2$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=-2$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

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

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

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

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

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

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

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

Pahekotia te 3x me x, ka 4x.

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

Tangohia te x mai i ngā taha e rua.

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

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

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

Me tāpiri te 2 ki ngā taha e rua.

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

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

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

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

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

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

$$a=1 b=2$$

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

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

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

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

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

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

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

$$x=-1 x=-2$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=-2$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

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

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

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

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

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

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

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

Pahekotia te 3x me x, ka 4x.

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

Tangohia te x mai i ngā taha e rua.

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

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

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

Me tāpiri te 2 ki ngā taha e rua.

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

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.

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

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

$$x=\frac{-3±\sqrt{3^{2}-4\times 2}}{2}$$

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\times 2}}{2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-3±\sqrt{9-8}}{2}$$

Tāpiri 9 ki te -8.

$$x=\frac{-3±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-3±1}{2}$$

Nā, me whakaoti te whārite x=\frac{-3±1}{2} ina he tāpiri te ±. Tāpiri -3 ki te 1.

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

Whakawehe -2 ki te 2.

$$x=-1$$

Nā, me whakaoti te whārite x=\frac{-3±1}{2} ina he tango te ±. Tango 1 mai i -3.

$$x=-\frac{4}{2}$$

Whakawehe -4 ki te 2.

$$x=-2$$

Kua oti te whārite te whakatau.

$$x=-1 x=-2$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=-2$$

Ngā Upane mō te Whakaoti i te Pūrua

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

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

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

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

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

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

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

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

Pahekotia te 3x me x, ka 4x.

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

Tangohia te x mai i ngā taha e rua.

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

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

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

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.

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

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

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

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

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

Tāpiri -2 ki te \frac{9}{4}.

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

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

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

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

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

Whakarūnātia.

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

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

$$x=-1 x=-2$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=-2$$