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

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

Whakaoti mō x

$$x=-2$$

$$x=-1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+4 ki te -2.

$$-2x-8+x-2=\left(x-2\right)\left(x+4\right)$$

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

$$-x-8-2=\left(x-2\right)\left(x+4\right)$$

Tangohia te 2 i te -8, ka -10.

$$-x-10=\left(x-2\right)\left(x+4\right)$$

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

$$-x-10=x^{2}+2x-8$$

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

$$-x-10-x^{2}=2x-8$$

Tangohia te 2x mai i ngā taha e rua.

$$-x-10-x^{2}-2x=-8$$

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

$$-3x-10-x^{2}=-8$$

Me tāpiri te 8 ki ngā taha e rua.

$$-3x-10-x^{2}+8=0$$

Tāpirihia te -10 ki te 8, ka -2.

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

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -1 mō a, -3 mō b, me -2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\left(-2\right)}}{2\left(-1\right)}$$

Whakareatia 4 ki te -2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8}}{2\left(-1\right)}$$

Tāpiri 9 ki te -8.

$$x=\frac{-\left(-3\right)±\sqrt{1}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-3\right)±1}{2\left(-1\right)}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±1}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{3±1}{-2}$$

Nā, me whakaoti te whārite x=\frac{3±1}{-2} ina he tāpiri te ±. Tāpiri 3 ki te 1.

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

Whakawehe 4 ki te -2.

$$x=-2$$

Nā, me whakaoti te whārite x=\frac{3±1}{-2} ina he tango te ±. Tango 1 mai i 3.

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

Whakawehe 2 ki te -2.

$$x=-1$$

Kua oti te whārite te whakatau.

$$x=-2 x=-1$$

Ngā Upane mō te Whakaoti i te Pūrua

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+4 ki te -2.

$$-2x-8+x-2=\left(x-2\right)\left(x+4\right)$$

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

$$-x-8-2=\left(x-2\right)\left(x+4\right)$$

Tangohia te 2 i te -8, ka -10.

$$-x-10=\left(x-2\right)\left(x+4\right)$$

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

$$-x-10=x^{2}+2x-8$$

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

$$-x-10-x^{2}=2x-8$$

Tangohia te 2x mai i ngā taha e rua.

$$-x-10-x^{2}-2x=-8$$

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

$$-3x-10-x^{2}=-8$$

Me tāpiri te 10 ki ngā taha e rua.

$$-3x-x^{2}=-8+10$$

Tāpirihia te -8 ki te 10, ka 2.

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

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.

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

Whakawehea ngā taha e rua ki te -1.

$$\frac{-x^{2}-3x}{-1}=\frac{2}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

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

Whakawehe -3 ki te -1.

$$x^{2}+3x=\frac{2}{-1}$$

Whakawehe 2 ki te -1.

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

Whakawehea te 3, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{2}. Nā, tāpiria te pūrua o te \frac{3}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+3x+\frac{9}{4}=-2+\frac{9}{4}$$

Tāpiri -2 ki te \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{1}{4}$$

Tauwehea x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$x+\frac{3}{2}=\frac{1}{2} x+\frac{3}{2}=-\frac{1}{2}$$

Me tango \frac{3}{2} mai i ngā taha e rua o te whārite.

$$x=-1 x=-2$$