Whakaoti i te 12(x+1)(x+3)/(x-1)^5=0

$$12 \frac { ( x + 1 ) ( x + 3 ) } { ( x - 1 ) ^ { 5 } } = 0$$

Whakaoti mō x

$$x=-3$$

$$x=-1$$

Ngā Upane Whakamahi Tauwehe

Whakawehea ngā taha e rua ki te 12. Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$\frac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^{5}}=0$$

Tē taea kia ōrite te tāupe x ki 1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te \left(x-1\right)^{5}.

$$\left(x+1\right)\left(x+3\right)=0$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te x+3 ka whakakotahi i ngā kupu rite.

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

Hei whakaoti i te whārite, whakatauwehea te x^{2}+4x+3 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.

$$a+b=4 ab=3$$

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

$$a=1 b=3$$

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

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

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

$$x=-1 x=-3$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakawehea ngā taha e rua ki te 12. Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$\frac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^{5}}=0$$

Tē taea kia ōrite te tāupe x ki 1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te \left(x-1\right)^{5}.

$$\left(x+1\right)\left(x+3\right)=0$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te x+3 ka whakakotahi i ngā kupu rite.

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

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.

$$a+b=4 ab=1\times 3=3$$

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

$$a=1 b=3$$

Tuhia anō te x^{2}+4x+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.

$$x=-1 x=-3$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakawehea ngā taha e rua ki te 12. Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$\frac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^{5}}=0$$

Tē taea kia ōrite te tāupe x ki 1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te \left(x-1\right)^{5}.

$$\left(x+1\right)\left(x+3\right)=0$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te x+3 ka whakakotahi i ngā kupu rite.

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

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

$$x=\frac{-4±\sqrt{4^{2}-4\times 3}}{2}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 3}}{2}$$

Whakareatia -4 ki te 3.

$$x=\frac{-4±\sqrt{16-12}}{2}$$

Tāpiri 16 ki te -12.

$$x=\frac{-4±\sqrt{4}}{2}$$

Tuhia te pūtakerua o te 4.

$$x=\frac{-4±2}{2}$$

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

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

Whakawehe -2 ki te 2.

$$x=-1$$

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

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

Whakawehe -6 ki te 2.

$$x=-3$$

Kua oti te whārite te whakatau.

$$x=-1 x=-3$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakawehea ngā taha e rua ki te 12. Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$\frac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^{5}}=0$$

Tē taea kia ōrite te tāupe x ki 1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te \left(x-1\right)^{5}.

$$\left(x+1\right)\left(x+3\right)=0$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te x+3 ka whakakotahi i ngā kupu rite.

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

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

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

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

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

Pūrua 2.

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

Tāpiri -3 ki te 4.

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

Tauwehea x^{2}+4x+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}.

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

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

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

Whakarūnātia.

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

Me tango 2 mai i ngā taha e rua o te whārite.

$$x=-1 x=-3$$