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

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

Whakaoti mō p

$$p=1$$

$$p=4$$

Ngā Upane Whakamahi Tauwehe

Tē taea kia ōrite te tāupe p ki tētahi o ngā uara -1,0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te p\left(p+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o p^{2}+p,p+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te p ki te p-6.

$$p+5=1-\left(p^{2}-6p\right)$$

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

$$p+5=1-p^{2}+6p$$

Tangohia te 1 mai i ngā taha e rua.

$$p+5-1=-p^{2}+6p$$

Tangohia te 1 i te 5, ka 4.

$$p+4=-p^{2}+6p$$

Me tāpiri te p^{2} ki ngā taha e rua.

$$p+4+p^{2}=6p$$

Tangohia te 6p mai i ngā taha e rua.

$$p+4+p^{2}-6p=0$$

Pahekotia te p me -6p, ka -5p.

$$-5p+4+p^{2}=0$$

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

$$p^{2}-5p+4=0$$

Hei whakaoti i te whārite, whakatauwehea te p^{2}-5p+4 mā te whakamahi i te tātai p^{2}+\left(a+b\right)p+ab=\left(p+a\right)\left(p+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=4$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 4.

$$-1,-4 -2,-2$$

Tātaihia te tapeke mō ia takirua.

$$-1-4=-5 -2-2=-4$$

Ko te otinga te takirua ka hoatu i te tapeke -5.

$$a=-4 b=-1$$

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

$$\left(p-4\right)\left(p-1\right)$$

Hei kimi otinga whārite, me whakaoti te p-4=0 me te p-1=0.

$$p=4 p=1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe p ki tētahi o ngā uara -1,0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te p\left(p+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o p^{2}+p,p+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te p ki te p-6.

$$p+5=1-\left(p^{2}-6p\right)$$

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

$$p+5=1-p^{2}+6p$$

Tangohia te 1 mai i ngā taha e rua.

$$p+5-1=-p^{2}+6p$$

Tangohia te 1 i te 5, ka 4.

$$p+4=-p^{2}+6p$$

Me tāpiri te p^{2} ki ngā taha e rua.

$$p+4+p^{2}=6p$$

Tangohia te 6p mai i ngā taha e rua.

$$p+4+p^{2}-6p=0$$

Pahekotia te p me -6p, ka -5p.

$$-5p+4+p^{2}=0$$

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

$$p^{2}-5p+4=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei p^{2}+ap+bp+4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=1\times 4=4$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 4.

$$-1,-4 -2,-2$$

Tātaihia te tapeke mō ia takirua.

$$-1-4=-5 -2-2=-4$$

Ko te otinga te takirua ka hoatu i te tapeke -5.

$$a=-4 b=-1$$

Tuhia anō te p^{2}-5p+4 hei \left(p^{2}-4p\right)+\left(-p+4\right).

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

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

$$p\left(p-4\right)-\left(p-4\right)$$

Whakatauwehea atu te kīanga pātahi p-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(p-4\right)\left(p-1\right)$$

Hei kimi otinga whārite, me whakaoti te p-4=0 me te p-1=0.

$$p=4 p=1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe p ki tētahi o ngā uara -1,0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te p\left(p+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o p^{2}+p,p+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te p ki te p-6.

$$p+5=1-\left(p^{2}-6p\right)$$

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

$$p+5=1-p^{2}+6p$$

Tangohia te 1 mai i ngā taha e rua.

$$p+5-1=-p^{2}+6p$$

Tangohia te 1 i te 5, ka 4.

$$p+4=-p^{2}+6p$$

Me tāpiri te p^{2} ki ngā taha e rua.

$$p+4+p^{2}=6p$$

Tangohia te 6p mai i ngā taha e rua.

$$p+4+p^{2}-6p=0$$

Pahekotia te p me -6p, ka -5p.

$$-5p+4+p^{2}=0$$

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.

$$p^{2}-5p+4=0$$

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

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

Pūrua -5.

$$p=\frac{-\left(-5\right)±\sqrt{25-4\times 4}}{2}$$

Whakareatia -4 ki te 4.

$$p=\frac{-\left(-5\right)±\sqrt{25-16}}{2}$$

Tāpiri 25 ki te -16.

$$p=\frac{-\left(-5\right)±\sqrt{9}}{2}$$

Tuhia te pūtakerua o te 9.

$$p=\frac{-\left(-5\right)±3}{2}$$

Ko te tauaro o -5 ko 5.

$$p=\frac{5±3}{2}$$

Nā, me whakaoti te whārite p=\frac{5±3}{2} ina he tāpiri te ±. Tāpiri 5 ki te 3.

$$p=\frac{8}{2}$$

Whakawehe 8 ki te 2.

$$p=4$$

Nā, me whakaoti te whārite p=\frac{5±3}{2} ina he tango te ±. Tango 3 mai i 5.

$$p=\frac{2}{2}$$

Whakawehe 2 ki te 2.

$$p=1$$

Kua oti te whārite te whakatau.

$$p=4 p=1$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe p ki tētahi o ngā uara -1,0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te p\left(p+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o p^{2}+p,p+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te p ki te p-6.

$$p+5=1-\left(p^{2}-6p\right)$$

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

$$p+5=1-p^{2}+6p$$

Me tāpiri te p^{2} ki ngā taha e rua.

$$p+5+p^{2}=1+6p$$

Tangohia te 6p mai i ngā taha e rua.

$$p+5+p^{2}-6p=1$$

Pahekotia te p me -6p, ka -5p.

$$-5p+5+p^{2}=1$$

Tangohia te 5 mai i ngā taha e rua.

$$-5p+p^{2}=1-5$$

Tangohia te 5 i te 1, ka -4.

$$-5p+p^{2}=-4$$

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.

$$p^{2}-5p=-4$$

Whakawehea te -5, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{2}. Nā, tāpiria te pūrua o te -\frac{5}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$p^{2}-5p+\left(-\frac{5}{2}\right)^{2}=-4+\left(-\frac{5}{2}\right)^{2}$$

Pūruatia -\frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$p^{2}-5p+\frac{25}{4}=-4+\frac{25}{4}$$

Tāpiri -4 ki te \frac{25}{4}.

$$p^{2}-5p+\frac{25}{4}=\frac{9}{4}$$

Tauwehea p^{2}-5p+\frac{25}{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}.

$$\left(p-\frac{5}{2}\right)^{2}=\frac{9}{4}$$

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

$$\sqrt{\left(p-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Whakarūnātia.

$$p-\frac{5}{2}=\frac{3}{2} p-\frac{5}{2}=-\frac{3}{2}$$

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$p=4 p=1$$