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

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

Whakaoti mō x

$$x=-4$$

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 2\left(x-2\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}-4,2-x,2x+4.

$$2\times 6-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Whakareatia te 2 ki te 6, ka 12.

$$12-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -4-2x ki te x+1 ka whakakotahi i ngā kupu rite.

$$12-\left(-6x-4-2x^{2}\right)=\left(x-2\right)x$$

Hei kimi i te tauaro o -6x-4-2x^{2}, kimihia te tauaro o ia taurangi.

$$12+6x+4+2x^{2}=\left(x-2\right)x$$

Tāpirihia te 12 ki te 4, ka 16.

$$16+6x+2x^{2}=\left(x-2\right)x$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-2 ki te x.

$$16+6x+2x^{2}=x^{2}-2x$$

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

$$16+6x+2x^{2}-x^{2}=-2x$$

Pahekotia te 2x^{2} me -x^{2}, ka x^{2}.

$$16+6x+x^{2}=-2x$$

Me tāpiri te 2x ki ngā taha e rua.

$$16+6x+x^{2}+2x=0$$

Pahekotia te 6x me 2x, ka 8x.

$$16+8x+x^{2}=0$$

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

$$x^{2}+8x+16=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}+8x+16 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.

$$a+b=8 ab=16$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 16.

$$1,16 2,8 4,4$$

Tātaihia te tapeke mō ia takirua.

$$1+16=17 2+8=10 4+4=8$$

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

$$a=4 b=4$$

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

$$\left(x+4\right)\left(x+4\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(x+4\right)^{2}$$

Hei kimi i te otinga whārite, whakaotia te x+4=0.

$$x=-4$$

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 2\left(x-2\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}-4,2-x,2x+4.

$$2\times 6-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Whakareatia te 2 ki te 6, ka 12.

$$12-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -4-2x ki te x+1 ka whakakotahi i ngā kupu rite.

$$12-\left(-6x-4-2x^{2}\right)=\left(x-2\right)x$$

Hei kimi i te tauaro o -6x-4-2x^{2}, kimihia te tauaro o ia taurangi.

$$12+6x+4+2x^{2}=\left(x-2\right)x$$

Tāpirihia te 12 ki te 4, ka 16.

$$16+6x+2x^{2}=\left(x-2\right)x$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-2 ki te x.

$$16+6x+2x^{2}=x^{2}-2x$$

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

$$16+6x+2x^{2}-x^{2}=-2x$$

Pahekotia te 2x^{2} me -x^{2}, ka x^{2}.

$$16+6x+x^{2}=-2x$$

Me tāpiri te 2x ki ngā taha e rua.

$$16+6x+x^{2}+2x=0$$

Pahekotia te 6x me 2x, ka 8x.

$$16+8x+x^{2}=0$$

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

$$x^{2}+8x+16=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx+16. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=8 ab=1\times 16=16$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 16.

$$1,16 2,8 4,4$$

Tātaihia te tapeke mō ia takirua.

$$1+16=17 2+8=10 4+4=8$$

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

$$a=4 b=4$$

Tuhia anō te x^{2}+8x+16 hei \left(x^{2}+4x\right)+\left(4x+16\right).

$$\left(x^{2}+4x\right)+\left(4x+16\right)$$

Tauwehea te x i te tuatahi me te 4 i te rōpū tuarua.

$$x\left(x+4\right)+4\left(x+4\right)$$

Whakatauwehea atu te kīanga pātahi x+4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x+4\right)\left(x+4\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(x+4\right)^{2}$$

Hei kimi i te otinga whārite, whakaotia te x+4=0.

$$x=-4$$

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 2\left(x-2\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}-4,2-x,2x+4.

$$2\times 6-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Whakareatia te 2 ki te 6, ka 12.

$$12-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -4-2x ki te x+1 ka whakakotahi i ngā kupu rite.

$$12-\left(-6x-4-2x^{2}\right)=\left(x-2\right)x$$

Hei kimi i te tauaro o -6x-4-2x^{2}, kimihia te tauaro o ia taurangi.

$$12+6x+4+2x^{2}=\left(x-2\right)x$$

Tāpirihia te 12 ki te 4, ka 16.

$$16+6x+2x^{2}=\left(x-2\right)x$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-2 ki te x.

$$16+6x+2x^{2}=x^{2}-2x$$

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

$$16+6x+2x^{2}-x^{2}=-2x$$

Pahekotia te 2x^{2} me -x^{2}, ka x^{2}.

$$16+6x+x^{2}=-2x$$

Me tāpiri te 2x ki ngā taha e rua.

$$16+6x+x^{2}+2x=0$$

Pahekotia te 6x me 2x, ka 8x.

$$16+8x+x^{2}=0$$

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.

$$x^{2}+8x+16=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 8 mō b, me 16 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 16}}{2}$$

Pūrua 8.

$$x=\frac{-8±\sqrt{64-4\times 16}}{2}$$

Whakareatia -4 ki te 16.

$$x=\frac{-8±\sqrt{64-64}}{2}$$

Tāpiri 64 ki te -64.

$$x=\frac{-8±\sqrt{0}}{2}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{8}{2}$$

Whakawehe -8 ki te 2.

$$x=-4$$

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 2\left(x-2\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}-4,2-x,2x+4.

$$2\times 6-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Whakareatia te 2 ki te 6, ka 12.

$$12-\left(-4-2x\right)\left(x+1\right)=\left(x-2\right)x$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -4-2x ki te x+1 ka whakakotahi i ngā kupu rite.

$$12-\left(-6x-4-2x^{2}\right)=\left(x-2\right)x$$

Hei kimi i te tauaro o -6x-4-2x^{2}, kimihia te tauaro o ia taurangi.

$$12+6x+4+2x^{2}=\left(x-2\right)x$$

Tāpirihia te 12 ki te 4, ka 16.

$$16+6x+2x^{2}=\left(x-2\right)x$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-2 ki te x.

$$16+6x+2x^{2}=x^{2}-2x$$

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

$$16+6x+2x^{2}-x^{2}=-2x$$

Pahekotia te 2x^{2} me -x^{2}, ka x^{2}.

$$16+6x+x^{2}=-2x$$

Me tāpiri te 2x ki ngā taha e rua.

$$16+6x+x^{2}+2x=0$$

Pahekotia te 6x me 2x, ka 8x.

$$16+8x+x^{2}=0$$

Tangohia te 16 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

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.

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

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.

$$x^{2}+8x+4^{2}=-16+4^{2}$$

Pūrua 4.

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

Tāpiri -16 ki te 16.

$$x^{2}+8x+16=0$$

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}.

$$\left(x+4\right)^{2}=0$$

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

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$x+4=0 x+4=0$$

Me tango 4 mai i ngā taha e rua o te whārite.

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

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

$$x=-4$$