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

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

Whakaoti mō x

$$x=-3$$

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-1,x-2,x^{2}-3x+2.

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

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

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

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

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

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

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

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

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

Hei kimi i te tauaro o x^{2}-2x+1, kimihia te tauaro o ia taurangi.

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

Pahekotia te x^{2} me -x^{2}, ka 0.

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

Pahekotia te -4x me 2x, ka -2x.

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

Tangohia te 1 i te 4, ka 3.

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

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

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

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

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

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

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

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=1 b=-3$$

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

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

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

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

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

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

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

$$x=1 x=-3$$

Tē taea kia ōrite te tāupe x ki 1.

$$x=-3$$

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-1,x-2,x^{2}-3x+2.

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

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

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

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

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

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

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

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

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

Hei kimi i te tauaro o x^{2}-2x+1, kimihia te tauaro o ia taurangi.

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

Pahekotia te x^{2} me -x^{2}, ka 0.

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

Pahekotia te -4x me 2x, ka -2x.

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

Tangohia te 1 i te 4, ka 3.

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

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

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

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.

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

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

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

Pūrua -2.

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

Whakareatia -4 ki te -1.

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

Whakareatia 4 ki te 3.

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

Tāpiri 4 ki te 12.

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

Tuhia te pūtakerua o te 16.

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

Ko te tauaro o -2 ko 2.

$$x=\frac{2±4}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{2±4}{-2}$$

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

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

Whakawehe 6 ki te -2.

$$x=-3$$

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

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

Whakawehe -2 ki te -2.

$$x=1$$

Kua oti te whārite te whakatau.

$$x=-3 x=1$$

Tē taea kia ōrite te tāupe x ki 1.

$$x=-3$$

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-1,x-2,x^{2}-3x+2.

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

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

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

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

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

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

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

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

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

Hei kimi i te tauaro o x^{2}-2x+1, kimihia te tauaro o ia taurangi.

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

Pahekotia te x^{2} me -x^{2}, ka 0.

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

Pahekotia te -4x me 2x, ka -2x.

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

Tangohia te 1 i te 4, ka 3.

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

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

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

Tangohia te 3 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

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.

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

Whakawehea ngā taha e rua ki te -1.

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

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

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

Whakawehe -2 ki te -1.

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

Whakawehe -3 ki te -1.

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

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

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

Pūrua 1.

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

Tāpiri 3 ki te 1.

$$x^{2}+2x+1=4$$

Tauwehea x^{2}+2x+1. 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}.

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

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

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

Whakarūnātia.

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

Me tango 1 mai i ngā taha e rua o te whārite.

$$x=1 x=-3$$

Tē taea kia ōrite te tāupe x ki 1.

$$x=-3$$