Whakaoti i te {l}{x-2/x+2=16/x^2-4}{=}

$$\left. \begin{array} { l } { \frac { x - 2 } { x + 2 } = \frac { 16 } { x ^ { 2 } - 4 } } \\ { = } \end{array} \right.$$

Whakaoti mō x

$$x=6$$

Ngā Upane Whakamahi Tauwehe

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,2 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+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+2,x^{2}-4.

$$\left(x-2\right)\left(x-2\right)=16$$

Whakareatia te x-2 ki te x-2, ka \left(x-2\right)^{2}.

$$\left(x-2\right)^{2}=16$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-2\right)^{2}.

$$x^{2}-4x+4=16$$

Tangohia te 16 mai i ngā taha e rua.

$$x^{2}-4x+4-16=0$$

Tangohia te 16 i te 4, ka -12.

$$x^{2}-4x-12=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}-4x-12 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.

$$a+b=-4 ab=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Tātaihia te tapeke mō ia takirua.

$$1-12=-11 2-6=-4 3-4=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -4.

$$a=-6 b=2$$

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

$$\left(x-6\right)\left(x+2\right)$$

Hei kimi otinga whārite, me whakaoti te x-6=0 me te x+2=0.

$$x=6 x=-2$$

Tē taea kia ōrite te tāupe x ki -2.

$$x=6$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,2 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+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+2,x^{2}-4.

$$\left(x-2\right)\left(x-2\right)=16$$

Whakareatia te x-2 ki te x-2, ka \left(x-2\right)^{2}.

$$\left(x-2\right)^{2}=16$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-2\right)^{2}.

$$x^{2}-4x+4=16$$

Tangohia te 16 mai i ngā taha e rua.

$$x^{2}-4x+4-16=0$$

Tangohia te 16 i te 4, ka -12.

$$x^{2}-4x-12=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=1\left(-12\right)=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Tātaihia te tapeke mō ia takirua.

$$1-12=-11 2-6=-4 3-4=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -4.

$$a=-6 b=2$$

Tuhia anō te x^{2}-4x-12 hei \left(x^{2}-6x\right)+\left(2x-12\right).

$$\left(x^{2}-6x\right)+\left(2x-12\right)$$

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

$$x\left(x-6\right)+2\left(x-6\right)$$

Whakatauwehea atu te kīanga pātahi x-6 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-6\right)\left(x+2\right)$$

Hei kimi otinga whārite, me whakaoti te x-6=0 me te x+2=0.

$$x=6 x=-2$$

Tē taea kia ōrite te tāupe x ki -2.

$$x=6$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,2 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+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+2,x^{2}-4.

$$\left(x-2\right)\left(x-2\right)=16$$

Whakareatia te x-2 ki te x-2, ka \left(x-2\right)^{2}.

$$\left(x-2\right)^{2}=16$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-2\right)^{2}.

$$x^{2}-4x+4=16$$

Tangohia te 16 mai i ngā taha e rua.

$$x^{2}-4x+4-16=0$$

Tangohia te 16 i te 4, ka -12.

$$x^{2}-4x-12=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -4 mō b, me -12 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-12\right)}}{2}$$

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-12\right)}}{2}$$

Whakareatia -4 ki te -12.

$$x=\frac{-\left(-4\right)±\sqrt{16+48}}{2}$$

Tāpiri 16 ki te 48.

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

Tuhia te pūtakerua o te 64.

$$x=\frac{-\left(-4\right)±8}{2}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±8}{2}$$

Nā, me whakaoti te whārite x=\frac{4±8}{2} ina he tāpiri te ±. Tāpiri 4 ki te 8.

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

Whakawehe 12 ki te 2.

$$x=6$$

Nā, me whakaoti te whārite x=\frac{4±8}{2} ina he tango te ±. Tango 8 mai i 4.

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

Whakawehe -4 ki te 2.

$$x=-2$$

Kua oti te whārite te whakatau.

$$x=6 x=-2$$

Tē taea kia ōrite te tāupe x ki -2.

$$x=6$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,2 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+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+2,x^{2}-4.

$$\left(x-2\right)\left(x-2\right)=16$$

Whakareatia te x-2 ki te x-2, ka \left(x-2\right)^{2}.

$$\left(x-2\right)^{2}=16$$

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

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{16}$$

Whakarūnātia.

$$x-2=4 x-2=-4$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$x=6 x=-2$$

Tē taea kia ōrite te tāupe x ki -2.

$$x=6$$