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

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

Whakaoti mō x (complex solution)

$$x=\sqrt{3}-1\approx 0.732050808$$

$$x=-\left(\sqrt{3}+1\right)\approx -2.732050808$$

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,x^{2}-4.

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

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

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

Pahekotia te x me 2x, ka 3x.

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

Tangohia te x mai i ngā taha e rua.

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

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

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

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.

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te -2.

$$x=\frac{-2±\sqrt{4+8}}{2}$$

Tāpiri 4 ki te 8.

$$x=\frac{-2±\sqrt{12}}{2}$$

Tuhia te pūtakerua o te 12.

$$x=\frac{-2±2\sqrt{3}}{2}$$

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

$$x=\frac{2\sqrt{3}-2}{2}$$

Whakawehe -2+2\sqrt{3} ki te 2.

$$x=\sqrt{3}-1$$

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

$$x=\frac{-2\sqrt{3}-2}{2}$$

Whakawehe -2-2\sqrt{3} ki te 2.

$$x=-\sqrt{3}-1$$

Kua oti te whārite te whakatau.

$$x=\sqrt{3}-1 x=-\sqrt{3}-1$$

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,x^{2}-4.

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

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

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

Pahekotia te x me 2x, ka 3x.

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

Tangohia te x mai i ngā taha e rua.

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

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

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

Me tāpiri te 2 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

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

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.

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

Whakawehea te 2, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te 1. Nā, tāpiria te pūrua o te 1 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūrua 1.

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

Tāpiri 2 ki te 1.

$$x^{2}+2x+1=3$$

Tauwehea x^{2}+2x+1. 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}.

$$\left(x+1\right)^{2}=3$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{3}$$

Whakarūnātia.

$$x+1=\sqrt{3} x+1=-\sqrt{3}$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$x=\sqrt{3}-1 x=-\sqrt{3}-1$$

Whakaoti mō x

$$x=\sqrt{3}-1\approx 0.732050808$$

$$x=-\sqrt{3}-1\approx -2.732050808$$

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,x^{2}-4.

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

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

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

Pahekotia te x me 2x, ka 3x.

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

Tangohia te x mai i ngā taha e rua.

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

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

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

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.

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te -2.

$$x=\frac{-2±\sqrt{4+8}}{2}$$

Tāpiri 4 ki te 8.

$$x=\frac{-2±\sqrt{12}}{2}$$

Tuhia te pūtakerua o te 12.

$$x=\frac{-2±2\sqrt{3}}{2}$$

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

$$x=\frac{2\sqrt{3}-2}{2}$$

Whakawehe -2+2\sqrt{3} ki te 2.

$$x=\sqrt{3}-1$$

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

$$x=\frac{-2\sqrt{3}-2}{2}$$

Whakawehe -2-2\sqrt{3} ki te 2.

$$x=-\sqrt{3}-1$$

Kua oti te whārite te whakatau.

$$x=\sqrt{3}-1 x=-\sqrt{3}-1$$

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,x^{2}-4.

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

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

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

Pahekotia te x me 2x, ka 3x.

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

Tangohia te x mai i ngā taha e rua.

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

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

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

Me tāpiri te 2 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

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

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.

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

Whakawehea te 2, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te 1. Nā, tāpiria te pūrua o te 1 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūrua 1.

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

Tāpiri 2 ki te 1.

$$x^{2}+2x+1=3$$

Tauwehea x^{2}+2x+1. 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}.

$$\left(x+1\right)^{2}=3$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{3}$$

Whakarūnātia.

$$x+1=\sqrt{3} x+1=-\sqrt{3}$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$x=\sqrt{3}-1 x=-\sqrt{3}-1$$