Whakaoti i te 1/(x-3)=2x+1/x^2-4

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

Whakaoti mō x

$$x = \frac{\sqrt{21} + 5}{2} \approx 4.791287847$$

$$x=\frac{5-\sqrt{21}}{2}\approx 0.208712153$$

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-3,x^{2}-4.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x-3 ki te 2x+1 ka whakakotahi i ngā kupu rite.

$$x^{2}-4=2x^{2}-5x-3$$

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

$$x^{2}-4-2x^{2}=-5x-3$$

Pahekotia te x^{2} me -2x^{2}, ka -x^{2}.

$$-x^{2}-4=-5x-3$$

Me tāpiri te 5x ki ngā taha e rua.

$$-x^{2}-4+5x=-3$$

Me tāpiri te 3 ki ngā taha e rua.

$$-x^{2}-4+5x+3=0$$

Tāpirihia te -4 ki te 3, ka -1.

$$-x^{2}-1+5x=0$$

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.

$$-x^{2}+5x-1=0$$

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

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-5±\sqrt{25+4\left(-1\right)}}{2\left(-1\right)}$$

Whakareatia 4 ki te -1.

$$x=\frac{-5±\sqrt{25-4}}{2\left(-1\right)}$$

Tāpiri 25 ki te -4.

$$x=\frac{-5±\sqrt{21}}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-5±\sqrt{21}}{-2}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{21}}{-2} ina he tāpiri te ±. Tāpiri -5 ki te \sqrt{21}.

$$x=\frac{\sqrt{21}-5}{-2}$$

Whakawehe -5+\sqrt{21} ki te -2.

$$x=\frac{5-\sqrt{21}}{2}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{21}}{-2} ina he tango te ±. Tango \sqrt{21} mai i -5.

$$x=\frac{-\sqrt{21}-5}{-2}$$

Whakawehe -5-\sqrt{21} ki te -2.

$$x=\frac{\sqrt{21}+5}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{5-\sqrt{21}}{2} x=\frac{\sqrt{21}+5}{2}$$

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-3,x^{2}-4.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x-3 ki te 2x+1 ka whakakotahi i ngā kupu rite.

$$x^{2}-4=2x^{2}-5x-3$$

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

$$x^{2}-4-2x^{2}=-5x-3$$

Pahekotia te x^{2} me -2x^{2}, ka -x^{2}.

$$-x^{2}-4=-5x-3$$

Me tāpiri te 5x ki ngā taha e rua.

$$-x^{2}-4+5x=-3$$

Me tāpiri te 4 ki ngā taha e rua.

$$-x^{2}+5x=-3+4$$

Tāpirihia te -3 ki te 4, ka 1.

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

Whakawehea ngā taha e rua ki te -1.

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

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

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

Whakawehe 5 ki te -1.

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

Whakawehe 1 ki te -1.

$$x^{2}-5x=-1$$

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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-1+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=-1+\frac{25}{4}$$

Tāpiri -1 ki te \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{21}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{21}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{\sqrt{21}}{2} x-\frac{5}{2}=-\frac{\sqrt{21}}{2}$$

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

$$x=\frac{\sqrt{21}+5}{2} x=\frac{5-\sqrt{21}}{2}$$