Whakaoti i te x/x^2-x-6=3/x^2-5x+6

$$\frac { x } { x ^ { 2 } - x - 6 } = \frac { 3 } { x ^ { 2 } - 5 x + 6 }$$

Whakaoti mō x

$$x=-1$$

$$x=6$$

Ngā Upane Whakamahi Tauwehe

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,2,3 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-3\right)\left(x-2\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}-x-6,x^{2}-5x+6.

$$\left(x-2\right)x=\left(x+2\right)\times 3$$

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+2 ki te 3.

$$x^{2}-2x=3x+6$$

Tangohia te 3x mai i ngā taha e rua.

$$x^{2}-2x-3x=6$$

Pahekotia te -2x me -3x, ka -5x.

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

Tangohia te 6 mai i ngā taha e rua.

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

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

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

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 -6.

$$1,-6 2,-3$$

Tātaihia te tapeke mō ia takirua.

$$1-6=-5 2-3=-1$$

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

$$a=-6 b=1$$

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

$$\left(x-6\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te x-6=0 me te x+1=0.

$$x=6 x=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,2,3 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-3\right)\left(x-2\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}-x-6,x^{2}-5x+6.

$$\left(x-2\right)x=\left(x+2\right)\times 3$$

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+2 ki te 3.

$$x^{2}-2x=3x+6$$

Tangohia te 3x mai i ngā taha e rua.

$$x^{2}-2x-3x=6$$

Pahekotia te -2x me -3x, ka -5x.

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

Tangohia te 6 mai i ngā taha e rua.

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=1\left(-6\right)=-6$$

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 -6.

$$1,-6 2,-3$$

Tātaihia te tapeke mō ia takirua.

$$1-6=-5 2-3=-1$$

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

$$a=-6 b=1$$

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

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

Whakatauwehea atu x i te x^{2}-6x.

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

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

$$\left(x-6\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te x-6=0 me te x+1=0.

$$x=6 x=-1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,2,3 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-3\right)\left(x-2\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}-x-6,x^{2}-5x+6.

$$\left(x-2\right)x=\left(x+2\right)\times 3$$

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+2 ki te 3.

$$x^{2}-2x=3x+6$$

Tangohia te 3x mai i ngā taha e rua.

$$x^{2}-2x-3x=6$$

Pahekotia te -2x me -3x, ka -5x.

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

Tangohia te 6 mai i ngā taha e rua.

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

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

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

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-6\right)}}{2}$$

Whakareatia -4 ki te -6.

$$x=\frac{-\left(-5\right)±\sqrt{25+24}}{2}$$

Tāpiri 25 ki te 24.

$$x=\frac{-\left(-5\right)±\sqrt{49}}{2}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-\left(-5\right)±7}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±7}{2}$$

Nā, me whakaoti te whārite x=\frac{5±7}{2} ina he tāpiri te ±. Tāpiri 5 ki te 7.

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

Whakawehe 12 ki te 2.

$$x=6$$

Nā, me whakaoti te whārite x=\frac{5±7}{2} ina he tango te ±. Tango 7 mai i 5.

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

Whakawehe -2 ki te 2.

$$x=-1$$

Kua oti te whārite te whakatau.

$$x=6 x=-1$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,2,3 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-3\right)\left(x-2\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x^{2}-x-6,x^{2}-5x+6.

$$\left(x-2\right)x=\left(x+2\right)\times 3$$

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+2 ki te 3.

$$x^{2}-2x=3x+6$$

Tangohia te 3x mai i ngā taha e rua.

$$x^{2}-2x-3x=6$$

Pahekotia te -2x me -3x, ka -5x.

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

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.

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

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

$$x^{2}-5x+\frac{25}{4}=6+\frac{25}{4}$$

Tāpiri 6 ki te \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{49}{4}$$

Tauwehea x^{2}-5x+\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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{49}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{7}{2} x-\frac{5}{2}=-\frac{7}{2}$$

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

$$x=6 x=-1$$