Whakaoti i te 1/2-x-1=1/x-2-5-x/3x^2-12

$$\frac { 1 } { 2 - x } - 1 = \frac { 1 } { x - 2 } - \frac { 5 - x } { 3 x ^ { 2 } - 12 }$$

Whakaoti mō x

$$x=\frac{\sqrt{109}-7}{6}\approx 0.573384418$$

$$x=\frac{-\sqrt{109}-7}{6}\approx -2.906717751$$

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 3\left(x-2\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2-x,x-2,3x^{2}-12.

$$-6-3x+3\left(x-2\right)\left(x+2\right)\left(-1\right)=3x+6-\left(5-x\right)$$

Whakareatia te 3 ki te -1, ka -3.

$$-6-3x-3\left(x-2\right)\left(x+2\right)=3x+6-\left(5-x\right)$$

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

$$-6-3x+\left(-3x+6\right)\left(x+2\right)=3x+6-\left(5-x\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -3x+6 ki te x+2 ka whakakotahi i ngā kupu rite.

$$-6-3x-3x^{2}+12=3x+6-\left(5-x\right)$$

Tāpirihia te -6 ki te 12, ka 6.

$$6-3x-3x^{2}=3x+6-\left(5-x\right)$$

Hei kimi i te tauaro o 5-x, kimihia te tauaro o ia taurangi.

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

Tangohia te 5 i te 6, ka 1.

$$6-3x-3x^{2}=3x+1+x$$

Pahekotia te 3x me x, ka 4x.

$$6-3x-3x^{2}=4x+1$$

Tangohia te 4x mai i ngā taha e rua.

$$6-3x-3x^{2}-4x=1$$

Pahekotia te -3x me -4x, ka -7x.

$$6-7x-3x^{2}=1$$

Tangohia te 1 mai i ngā taha e rua.

$$6-7x-3x^{2}-1=0$$

Tangohia te 1 i te 6, ka 5.

$$5-7x-3x^{2}=0$$

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.

$$-3x^{2}-7x+5=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -3 mō a, -7 mō b, me 5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-3\right)\times 5}}{2\left(-3\right)}$$

Pūrua -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-3\right)\times 5}}{2\left(-3\right)}$$

Whakareatia -4 ki te -3.

$$x=\frac{-\left(-7\right)±\sqrt{49+12\times 5}}{2\left(-3\right)}$$

Whakareatia 12 ki te 5.

$$x=\frac{-\left(-7\right)±\sqrt{49+60}}{2\left(-3\right)}$$

Tāpiri 49 ki te 60.

$$x=\frac{-\left(-7\right)±\sqrt{109}}{2\left(-3\right)}$$

Ko te tauaro o -7 ko 7.

$$x=\frac{7±\sqrt{109}}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{7±\sqrt{109}}{-6}$$

Nā, me whakaoti te whārite x=\frac{7±\sqrt{109}}{-6} ina he tāpiri te ±. Tāpiri 7 ki te \sqrt{109}.

$$x=\frac{\sqrt{109}+7}{-6}$$

Whakawehe 7+\sqrt{109} ki te -6.

$$x=\frac{-\sqrt{109}-7}{6}$$

Nā, me whakaoti te whārite x=\frac{7±\sqrt{109}}{-6} ina he tango te ±. Tango \sqrt{109} mai i 7.

$$x=\frac{7-\sqrt{109}}{-6}$$

Whakawehe 7-\sqrt{109} ki te -6.

$$x=\frac{\sqrt{109}-7}{6}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{109}-7}{6} x=\frac{\sqrt{109}-7}{6}$$

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 3\left(x-2\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2-x,x-2,3x^{2}-12.

$$-6-3x+3\left(x-2\right)\left(x+2\right)\left(-1\right)=3x+6-\left(5-x\right)$$

Whakareatia te 3 ki te -1, ka -3.

$$-6-3x-3\left(x-2\right)\left(x+2\right)=3x+6-\left(5-x\right)$$

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

$$-6-3x+\left(-3x+6\right)\left(x+2\right)=3x+6-\left(5-x\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -3x+6 ki te x+2 ka whakakotahi i ngā kupu rite.

$$-6-3x-3x^{2}+12=3x+6-\left(5-x\right)$$

Tāpirihia te -6 ki te 12, ka 6.

$$6-3x-3x^{2}=3x+6-\left(5-x\right)$$

Hei kimi i te tauaro o 5-x, kimihia te tauaro o ia taurangi.

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

Tangohia te 5 i te 6, ka 1.

$$6-3x-3x^{2}=3x+1+x$$

Pahekotia te 3x me x, ka 4x.

$$6-3x-3x^{2}=4x+1$$

Tangohia te 4x mai i ngā taha e rua.

$$6-3x-3x^{2}-4x=1$$

Pahekotia te -3x me -4x, ka -7x.

$$6-7x-3x^{2}=1$$

Tangohia te 6 mai i ngā taha e rua.

$$-7x-3x^{2}=1-6$$

Tangohia te 6 i te 1, ka -5.

$$-7x-3x^{2}=-5$$

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.

$$-3x^{2}-7x=-5$$

Whakawehea ngā taha e rua ki te -3.

$$\frac{-3x^{2}-7x}{-3}=-\frac{5}{-3}$$

Mā te whakawehe ki te -3 ka wetekia te whakareanga ki te -3.

$$x^{2}+\left(-\frac{7}{-3}\right)x=-\frac{5}{-3}$$

Whakawehe -7 ki te -3.

$$x^{2}+\frac{7}{3}x=-\frac{5}{-3}$$

Whakawehe -5 ki te -3.

$$x^{2}+\frac{7}{3}x=\frac{5}{3}$$

Whakawehea te \frac{7}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{7}{6}. Nā, tāpiria te pūrua o te \frac{7}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=\frac{5}{3}+\left(\frac{7}{6}\right)^{2}$$

Pūruatia \frac{7}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{5}{3}+\frac{49}{36}$$

Tāpiri \frac{5}{3} ki te \frac{49}{36} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{109}{36}$$

Tauwehea x^{2}+\frac{7}{3}x+\frac{49}{36}. 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}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{109}{36}$$

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

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{109}{36}}$$

Whakarūnātia.

$$x+\frac{7}{6}=\frac{\sqrt{109}}{6} x+\frac{7}{6}=-\frac{\sqrt{109}}{6}$$

Me tango \frac{7}{6} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{109}-7}{6} x=\frac{-\sqrt{109}-7}{6}$$