Whakaoti i te 10/x^2-2x-8+5/x+2+1=0

$$\frac{ 10 }{ { x }^{ 2 } -2x-8 } + \frac{ 5 }{ x+2 } +1=0$$

Whakaoti mō x

$$x=-6$$

$$x=3$$

Ngā Upane Whakamahi Tauwehe

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,4 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-4\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}-2x-8,x+2.

$$10+\left(x-4\right)\times 5+\left(x-4\right)\left(x+2\right)=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-4 ki te 5.

$$10+5x-20+\left(x-4\right)\left(x+2\right)=0$$

Tangohia te 20 i te 10, ka -10.

$$-10+5x+\left(x-4\right)\left(x+2\right)=0$$

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

$$-10+5x+x^{2}-2x-8=0$$

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

$$-10+3x+x^{2}-8=0$$

Tangohia te 8 i te -10, ka -18.

$$-18+3x+x^{2}=0$$

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

$$x^{2}+3x-18=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}+3x-18 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.

$$a+b=3 ab=-18$$

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 -18.

$$-1,18 -2,9 -3,6$$

Tātaihia te tapeke mō ia takirua.

$$-1+18=17 -2+9=7 -3+6=3$$

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

$$a=-3 b=6$$

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

$$\left(x-3\right)\left(x+6\right)$$

Hei kimi otinga whārite, me whakaoti te x-3=0 me te x+6=0.

$$x=3 x=-6$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,4 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-4\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}-2x-8,x+2.

$$10+\left(x-4\right)\times 5+\left(x-4\right)\left(x+2\right)=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-4 ki te 5.

$$10+5x-20+\left(x-4\right)\left(x+2\right)=0$$

Tangohia te 20 i te 10, ka -10.

$$-10+5x+\left(x-4\right)\left(x+2\right)=0$$

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

$$-10+5x+x^{2}-2x-8=0$$

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

$$-10+3x+x^{2}-8=0$$

Tangohia te 8 i te -10, ka -18.

$$-18+3x+x^{2}=0$$

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

$$x^{2}+3x-18=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-18. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=3 ab=1\left(-18\right)=-18$$

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 -18.

$$-1,18 -2,9 -3,6$$

Tātaihia te tapeke mō ia takirua.

$$-1+18=17 -2+9=7 -3+6=3$$

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

$$a=-3 b=6$$

Tuhia anō te x^{2}+3x-18 hei \left(x^{2}-3x\right)+\left(6x-18\right).

$$\left(x^{2}-3x\right)+\left(6x-18\right)$$

Tauwehea te x i te tuatahi me te 6 i te rōpū tuarua.

$$x\left(x-3\right)+6\left(x-3\right)$$

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

$$\left(x-3\right)\left(x+6\right)$$

Hei kimi otinga whārite, me whakaoti te x-3=0 me te x+6=0.

$$x=3 x=-6$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,4 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-4\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}-2x-8,x+2.

$$10+\left(x-4\right)\times 5+\left(x-4\right)\left(x+2\right)=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-4 ki te 5.

$$10+5x-20+\left(x-4\right)\left(x+2\right)=0$$

Tangohia te 20 i te 10, ka -10.

$$-10+5x+\left(x-4\right)\left(x+2\right)=0$$

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

$$-10+5x+x^{2}-2x-8=0$$

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

$$-10+3x+x^{2}-8=0$$

Tangohia te 8 i te -10, ka -18.

$$-18+3x+x^{2}=0$$

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.

$$x^{2}+3x-18=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 3 mō b, me -18 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-18\right)}}{2}$$

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\left(-18\right)}}{2}$$

Whakareatia -4 ki te -18.

$$x=\frac{-3±\sqrt{9+72}}{2}$$

Tāpiri 9 ki te 72.

$$x=\frac{-3±\sqrt{81}}{2}$$

Tuhia te pūtakerua o te 81.

$$x=\frac{-3±9}{2}$$

Nā, me whakaoti te whārite x=\frac{-3±9}{2} ina he tāpiri te ±. Tāpiri -3 ki te 9.

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

Whakawehe 6 ki te 2.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{-3±9}{2} ina he tango te ±. Tango 9 mai i -3.

$$x=-\frac{12}{2}$$

Whakawehe -12 ki te 2.

$$x=-6$$

Kua oti te whārite te whakatau.

$$x=3 x=-6$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,4 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-4\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}-2x-8,x+2.

$$10+\left(x-4\right)\times 5+\left(x-4\right)\left(x+2\right)=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-4 ki te 5.

$$10+5x-20+\left(x-4\right)\left(x+2\right)=0$$

Tangohia te 20 i te 10, ka -10.

$$-10+5x+\left(x-4\right)\left(x+2\right)=0$$

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

$$-10+5x+x^{2}-2x-8=0$$

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

$$-10+3x+x^{2}-8=0$$

Tangohia te 8 i te -10, ka -18.

$$-18+3x+x^{2}=0$$

Me tāpiri te 18 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$3x+x^{2}=18$$

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.

$$x^{2}+3x=18$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=18+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=18+\frac{9}{4}$$

Tāpiri 18 ki te \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{81}{4}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{81}{4}$$

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

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Whakarūnātia.

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

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

$$x=3 x=-6$$