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

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

Whakaoti mō x

$$x=5$$

Ngā Upane Whakamahi Tauwehe

Tē taea kia ōrite te tāupe x ki 1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x-1, arā, te tauraro pātahi he tino iti rawa te kitea o x-1,1-x.

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

Me tāpiri te 5 ki ngā taha e rua.

$$x^{2}-6x+5=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}-6x+5 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.

$$a+b=-6 ab=5$$

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

$$a=-5 b=-1$$

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

$$\left(x-5\right)\left(x-1\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x-1=0.

$$x=5 x=1$$

Tē taea kia ōrite te tāupe x ki 1.

$$x=5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe x ki 1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x-1, arā, te tauraro pātahi he tino iti rawa te kitea o x-1,1-x.

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

Me tāpiri te 5 ki ngā taha e rua.

$$x^{2}-6x+5=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx+5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-6 ab=1\times 5=5$$

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

$$a=-5 b=-1$$

Tuhia anō te x^{2}-6x+5 hei \left(x^{2}-5x\right)+\left(-x+5\right).

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

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

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

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

$$\left(x-5\right)\left(x-1\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x-1=0.

$$x=5 x=1$$

Tē taea kia ōrite te tāupe x ki 1.

$$x=5$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x-1, arā, te tauraro pātahi he tino iti rawa te kitea o x-1,1-x.

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

Me tāpiri te 5 ki ngā taha e rua.

$$x^{2}-6x+5=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -6 mō b, me 5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5}}{2}$$

Pūrua -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 5}}{2}$$

Whakareatia -4 ki te 5.

$$x=\frac{-\left(-6\right)±\sqrt{36-20}}{2}$$

Tāpiri 36 ki te -20.

$$x=\frac{-\left(-6\right)±\sqrt{16}}{2}$$

Tuhia te pūtakerua o te 16.

$$x=\frac{-\left(-6\right)±4}{2}$$

Ko te tauaro o -6 ko 6.

$$x=\frac{6±4}{2}$$

Nā, me whakaoti te whārite x=\frac{6±4}{2} ina he tāpiri te ±. Tāpiri 6 ki te 4.

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

Whakawehe 10 ki te 2.

$$x=5$$

Nā, me whakaoti te whārite x=\frac{6±4}{2} ina he tango te ±. Tango 4 mai i 6.

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

Whakawehe 2 ki te 2.

$$x=1$$

Kua oti te whārite te whakatau.

$$x=5 x=1$$

Tē taea kia ōrite te tāupe x ki 1.

$$x=5$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki 1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x-1, arā, te tauraro pātahi he tino iti rawa te kitea o x-1,1-x.

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

Whakawehea te -6, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -3. Nā, tāpiria te pūrua o te -3 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūrua -3.

$$x^{2}-6x+9=-5+9$$

Tāpiri -5 ki te 9.

$$x^{2}-6x+9=4$$

Tauwehea x^{2}-6x+9. 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}.

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

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

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

Whakarūnātia.

$$x-3=2 x-3=-2$$

Me tāpiri 3 ki ngā taha e rua o te whārite.

$$x=5 x=1$$

Tē taea kia ōrite te tāupe x ki 1.

$$x=5$$