Whakaoti i te 6x/x^2-1-5/1-x=x+4/x+1

$$\frac { 6 x } { x ^ { 2 } - 1 } - \frac { 5 } { 1 - x } = \frac { x + 4 } { x + 1 }$$

Whakaoti mō x

$$x=9$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

$$6x-\left(-\left(1+x\right)\times 5\right)=\left(x-1\right)\left(x+4\right)$$

Whakareatia te -1 ki te 5, ka -5.

$$6x-\left(-5\left(1+x\right)\right)=\left(x-1\right)\left(x+4\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -5 ki te 1+x.

$$6x-\left(-5-5x\right)=\left(x-1\right)\left(x+4\right)$$

Hei kimi i te tauaro o -5-5x, kimihia te tauaro o ia taurangi.

$$6x+5+5x=\left(x-1\right)\left(x+4\right)$$

Pahekotia te 6x me 5x, ka 11x.

$$11x+5=\left(x-1\right)\left(x+4\right)$$

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

$$11x+5=x^{2}+3x-4$$

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

$$11x+5-x^{2}=3x-4$$

Tangohia te 3x mai i ngā taha e rua.

$$11x+5-x^{2}-3x=-4$$

Pahekotia te 11x me -3x, ka 8x.

$$8x+5-x^{2}=-4$$

Me tāpiri te 4 ki ngā taha e rua.

$$8x+5-x^{2}+4=0$$

Tāpirihia te 5 ki te 4, ka 9.

$$8x+9-x^{2}=0$$

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

$$-x^{2}+8x+9=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -x^{2}+ax+bx+9. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=8 ab=-9=-9$$

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 -9.

$$-1,9 -3,3$$

Tātaihia te tapeke mō ia takirua.

$$-1+9=8 -3+3=0$$

Ko te otinga te takirua ka hoatu i te tapeke 8.

$$a=9 b=-1$$

Tuhia anō te -x^{2}+8x+9 hei \left(-x^{2}+9x\right)+\left(-x+9\right).

$$\left(-x^{2}+9x\right)+\left(-x+9\right)$$

Tauwehea te -x i te tuatahi me te -1 i te rōpū tuarua.

$$-x\left(x-9\right)-\left(x-9\right)$$

Whakatauwehea atu te kīanga pātahi x-9 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-9\right)\left(-x-1\right)$$

Hei kimi otinga whārite, me whakaoti te x-9=0 me te -x-1=0.

$$x=9 x=-1$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=9$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$6x-\left(-\left(1+x\right)\times 5\right)=\left(x-1\right)\left(x+4\right)$$

Whakareatia te -1 ki te 5, ka -5.

$$6x-\left(-5\left(1+x\right)\right)=\left(x-1\right)\left(x+4\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -5 ki te 1+x.

$$6x-\left(-5-5x\right)=\left(x-1\right)\left(x+4\right)$$

Hei kimi i te tauaro o -5-5x, kimihia te tauaro o ia taurangi.

$$6x+5+5x=\left(x-1\right)\left(x+4\right)$$

Pahekotia te 6x me 5x, ka 11x.

$$11x+5=\left(x-1\right)\left(x+4\right)$$

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

$$11x+5=x^{2}+3x-4$$

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

$$11x+5-x^{2}=3x-4$$

Tangohia te 3x mai i ngā taha e rua.

$$11x+5-x^{2}-3x=-4$$

Pahekotia te 11x me -3x, ka 8x.

$$8x+5-x^{2}=-4$$

Me tāpiri te 4 ki ngā taha e rua.

$$8x+5-x^{2}+4=0$$

Tāpirihia te 5 ki te 4, ka 9.

$$8x+9-x^{2}=0$$

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.

$$-x^{2}+8x+9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -1 mō a, 8 mō b, me 9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Pūrua 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-8±\sqrt{64+4\times 9}}{2\left(-1\right)}$$

Whakareatia 4 ki te 9.

$$x=\frac{-8±\sqrt{64+36}}{2\left(-1\right)}$$

Tāpiri 64 ki te 36.

$$x=\frac{-8±\sqrt{100}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 100.

$$x=\frac{-8±10}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-8±10}{-2}$$

Nā, me whakaoti te whārite x=\frac{-8±10}{-2} ina he tāpiri te ±. Tāpiri -8 ki te 10.

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

Whakawehe 2 ki te -2.

$$x=-1$$

Nā, me whakaoti te whārite x=\frac{-8±10}{-2} ina he tango te ±. Tango 10 mai i -8.

$$x=-\frac{18}{-2}$$

Whakawehe -18 ki te -2.

$$x=9$$

Kua oti te whārite te whakatau.

$$x=-1 x=9$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=9$$

Ngā Upane mō te Whakaoti i te Pūrua

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

$$6x-\left(-\left(1+x\right)\times 5\right)=\left(x-1\right)\left(x+4\right)$$

Whakareatia te -1 ki te 5, ka -5.

$$6x-\left(-5\left(1+x\right)\right)=\left(x-1\right)\left(x+4\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -5 ki te 1+x.

$$6x-\left(-5-5x\right)=\left(x-1\right)\left(x+4\right)$$

Hei kimi i te tauaro o -5-5x, kimihia te tauaro o ia taurangi.

$$6x+5+5x=\left(x-1\right)\left(x+4\right)$$

Pahekotia te 6x me 5x, ka 11x.

$$11x+5=\left(x-1\right)\left(x+4\right)$$

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

$$11x+5=x^{2}+3x-4$$

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

$$11x+5-x^{2}=3x-4$$

Tangohia te 3x mai i ngā taha e rua.

$$11x+5-x^{2}-3x=-4$$

Pahekotia te 11x me -3x, ka 8x.

$$8x+5-x^{2}=-4$$

Tangohia te 5 mai i ngā taha e rua.

$$8x-x^{2}=-4-5$$

Tangohia te 5 i te -4, ka -9.

$$8x-x^{2}=-9$$

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.

$$-x^{2}+8x=-9$$

Whakawehea ngā taha e rua ki te -1.

$$\frac{-x^{2}+8x}{-1}=-\frac{9}{-1}$$

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

$$x^{2}+\frac{8}{-1}x=-\frac{9}{-1}$$

Whakawehe 8 ki te -1.

$$x^{2}-8x=-\frac{9}{-1}$$

Whakawehe -9 ki te -1.

$$x^{2}-8x=9$$

Whakawehea te -8, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -4. Nā, tāpiria te pūrua o te -4 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-8x+\left(-4\right)^{2}=9+\left(-4\right)^{2}$$

Pūrua -4.

$$x^{2}-8x+16=9+16$$

Tāpiri 9 ki te 16.

$$x^{2}-8x+16=25$$

Tauwehea x^{2}-8x+16. 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}.

$$\left(x-4\right)^{2}=25$$

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

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{25}$$

Whakarūnātia.

$$x-4=5 x-4=-5$$

Me tāpiri 4 ki ngā taha e rua o te whārite.

$$x=9 x=-1$$

Tē taea kia ōrite te tāupe x ki -1.

$$x=9$$