Whakaoti i te 1-x/x^2+5x+6-3/x+2+2left(x+1right)/x+3=1

$$\frac{ 1-x }{ { x }^{ 2 } +5x+6 } - \frac{ 3 }{ x+2 } + \frac{ 2 \left( x+1 \right) }{ x+3 } = 1$$

Whakaoti mō x

$$x=5$$

Ngā Upane Whakamahi Tauwehe

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -3,-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+3\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}+5x+6,x+2,x+3.

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

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

$$1-x-\left(3x+9\right)+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

Hei kimi i te tauaro o 3x+9, kimihia te tauaro o ia taurangi.

$$1-x-3x-9+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

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

$$1-4x-9+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

Tangohia te 9 i te 1, ka -8.

$$-8-4x+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

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

$$-8-4x+\left(2x+4\right)\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 2x+4 ki te x+1 ka whakakotahi i ngā kupu rite.

$$-8-4x+2x^{2}+6x+4=\left(x+2\right)\left(x+3\right)$$

Pahekotia te -4x me 6x, ka 2x.

$$-8+2x+2x^{2}+4=\left(x+2\right)\left(x+3\right)$$

Tāpirihia te -8 ki te 4, ka -4.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+2 ki te x+3 ka whakakotahi i ngā kupu rite.

$$-4+2x+2x^{2}=x^{2}+5x+6$$

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

$$-4+2x+2x^{2}-x^{2}=5x+6$$

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

$$-4+2x+x^{2}=5x+6$$

Tangohia te 5x mai i ngā taha e rua.

$$-4+2x+x^{2}-5x=6$$

Pahekotia te 2x me -5x, ka -3x.

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

Tangohia te 6 mai i ngā taha e rua.

$$-4-3x+x^{2}-6=0$$

Tangohia te 6 i te -4, ka -10.

$$-10-3x+x^{2}=0$$

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

$$x^{2}-3x-10=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}-3x-10 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.

$$a+b=-3 ab=-10$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -10.

$$1,-10 2,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-10=-9 2-5=-3$$

Ko te otinga te takirua ka hoatu i te tapeke -3.

$$a=-5 b=2$$

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

$$\left(x-5\right)\left(x+2\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x+2=0.

$$x=5 x=-2$$

Tē taea kia ōrite te tāupe x ki -2.

$$x=5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -3,-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+3\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}+5x+6,x+2,x+3.

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

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

$$1-x-\left(3x+9\right)+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

Hei kimi i te tauaro o 3x+9, kimihia te tauaro o ia taurangi.

$$1-x-3x-9+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

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

$$1-4x-9+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

Tangohia te 9 i te 1, ka -8.

$$-8-4x+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

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

$$-8-4x+\left(2x+4\right)\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 2x+4 ki te x+1 ka whakakotahi i ngā kupu rite.

$$-8-4x+2x^{2}+6x+4=\left(x+2\right)\left(x+3\right)$$

Pahekotia te -4x me 6x, ka 2x.

$$-8+2x+2x^{2}+4=\left(x+2\right)\left(x+3\right)$$

Tāpirihia te -8 ki te 4, ka -4.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+2 ki te x+3 ka whakakotahi i ngā kupu rite.

$$-4+2x+2x^{2}=x^{2}+5x+6$$

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

$$-4+2x+2x^{2}-x^{2}=5x+6$$

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

$$-4+2x+x^{2}=5x+6$$

Tangohia te 5x mai i ngā taha e rua.

$$-4+2x+x^{2}-5x=6$$

Pahekotia te 2x me -5x, ka -3x.

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

Tangohia te 6 mai i ngā taha e rua.

$$-4-3x+x^{2}-6=0$$

Tangohia te 6 i te -4, ka -10.

$$-10-3x+x^{2}=0$$

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

$$x^{2}-3x-10=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-10. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=1\left(-10\right)=-10$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -10.

$$1,-10 2,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-10=-9 2-5=-3$$

Ko te otinga te takirua ka hoatu i te tapeke -3.

$$a=-5 b=2$$

Tuhia anō te x^{2}-3x-10 hei \left(x^{2}-5x\right)+\left(2x-10\right).

$$\left(x^{2}-5x\right)+\left(2x-10\right)$$

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

$$x\left(x-5\right)+2\left(x-5\right)$$

Whakatauwehea atu te kīanga pātahi x-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-5\right)\left(x+2\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x+2=0.

$$x=5 x=-2$$

Tē taea kia ōrite te tāupe x ki -2.

$$x=5$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -3,-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+3\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}+5x+6,x+2,x+3.

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

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

$$1-x-\left(3x+9\right)+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

Hei kimi i te tauaro o 3x+9, kimihia te tauaro o ia taurangi.

$$1-x-3x-9+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

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

$$1-4x-9+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

Tangohia te 9 i te 1, ka -8.

$$-8-4x+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

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

$$-8-4x+\left(2x+4\right)\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 2x+4 ki te x+1 ka whakakotahi i ngā kupu rite.

$$-8-4x+2x^{2}+6x+4=\left(x+2\right)\left(x+3\right)$$

Pahekotia te -4x me 6x, ka 2x.

$$-8+2x+2x^{2}+4=\left(x+2\right)\left(x+3\right)$$

Tāpirihia te -8 ki te 4, ka -4.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+2 ki te x+3 ka whakakotahi i ngā kupu rite.

$$-4+2x+2x^{2}=x^{2}+5x+6$$

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

$$-4+2x+2x^{2}-x^{2}=5x+6$$

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

$$-4+2x+x^{2}=5x+6$$

Tangohia te 5x mai i ngā taha e rua.

$$-4+2x+x^{2}-5x=6$$

Pahekotia te 2x me -5x, ka -3x.

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

Tangohia te 6 mai i ngā taha e rua.

$$-4-3x+x^{2}-6=0$$

Tangohia te 6 i te -4, ka -10.

$$-10-3x+x^{2}=0$$

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.

$$x^{2}-3x-10=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -3 mō b, me -10 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-10\right)}}{2}$$

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-10\right)}}{2}$$

Whakareatia -4 ki te -10.

$$x=\frac{-\left(-3\right)±\sqrt{9+40}}{2}$$

Tāpiri 9 ki te 40.

$$x=\frac{-\left(-3\right)±\sqrt{49}}{2}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-\left(-3\right)±7}{2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±7}{2}$$

Nā, me whakaoti te whārite x=\frac{3±7}{2} ina he tāpiri te ±. Tāpiri 3 ki te 7.

$$x=\frac{10}{2}$$

Whakawehe 10 ki te 2.

$$x=5$$

Nā, me whakaoti te whārite x=\frac{3±7}{2} ina he tango te ±. Tango 7 mai i 3.

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

Whakawehe -4 ki te 2.

$$x=-2$$

Kua oti te whārite te whakatau.

$$x=5 x=-2$$

Tē taea kia ōrite te tāupe x ki -2.

$$x=5$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -3,-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+3\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}+5x+6,x+2,x+3.

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

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

$$1-x-\left(3x+9\right)+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

Hei kimi i te tauaro o 3x+9, kimihia te tauaro o ia taurangi.

$$1-x-3x-9+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

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

$$1-4x-9+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

Tangohia te 9 i te 1, ka -8.

$$-8-4x+\left(x+2\right)\times 2\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

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

$$-8-4x+\left(2x+4\right)\left(x+1\right)=\left(x+2\right)\left(x+3\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 2x+4 ki te x+1 ka whakakotahi i ngā kupu rite.

$$-8-4x+2x^{2}+6x+4=\left(x+2\right)\left(x+3\right)$$

Pahekotia te -4x me 6x, ka 2x.

$$-8+2x+2x^{2}+4=\left(x+2\right)\left(x+3\right)$$

Tāpirihia te -8 ki te 4, ka -4.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+2 ki te x+3 ka whakakotahi i ngā kupu rite.

$$-4+2x+2x^{2}=x^{2}+5x+6$$

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

$$-4+2x+2x^{2}-x^{2}=5x+6$$

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

$$-4+2x+x^{2}=5x+6$$

Tangohia te 5x mai i ngā taha e rua.

$$-4+2x+x^{2}-5x=6$$

Pahekotia te 2x me -5x, ka -3x.

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

Me tāpiri te 4 ki ngā taha e rua.

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

Tāpirihia te 6 ki te 4, ka 10.

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

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.

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

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.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=10+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=10+\frac{9}{4}$$

Tāpiri 10 ki te \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{49}{4}$$

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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{49}{4}$$

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

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Whakarūnātia.

$$x-\frac{3}{2}=\frac{7}{2} x-\frac{3}{2}=-\frac{7}{2}$$

Me tāpiri \frac{3}{2} ki ngā taha e rua o te whārite.

$$x=5 x=-2$$

Tē taea kia ōrite te tāupe x ki -2.

$$x=5$$